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mathematics Mathematics is a field of study that discovers and organizes methods, Mathematical theory, theories and theorems that are developed and Mathematical proof, proved for the needs of empirical sciences and mathematics itself. There are many ar ...
, the following inequality is known as Titu's lemma, Bergström's inequality, Engel's form or Sedrakyan's inequality, respectively, referring to the article ''About the applications of one useful inequality'' of Nairi Sedrakyan published in 1997, to the book ''Problem-solving strategies'' of Arthur Engel published in 1998 and to the book ''Mathematical Olympiad Treasures'' of Titu Andreescu published in 2003. It is a direct consequence of Cauchy–Bunyakovsky–Schwarz inequality. Nevertheless, in his article (1997) Sedrakyan has noticed that written in this form this
inequality Inequality may refer to: * Inequality (mathematics), a relation between two quantities when they are different. * Economic inequality, difference in economic well-being between population groups ** Income inequality, an unequal distribution of i ...
can be used as a
proof Proof most often refers to: * Proof (truth), argument or sufficient evidence for the truth of a proposition * Alcohol proof, a measure of an alcoholic drink's strength Proof may also refer to: Mathematics and formal logic * Formal proof, a co ...
technique and it has very useful new applications. In the book ''Algebraic Inequalities'' (Sedrakyan) several generalizations of this inequality are provided.


Statement of the inequality

For any
real number In mathematics, a real number is a number that can be used to measure a continuous one- dimensional quantity such as a duration or temperature. Here, ''continuous'' means that pairs of values can have arbitrarily small differences. Every re ...
s a_1, a_2, a_3, \ldots, a_n and positive reals b_1, b_2, b_3,\ldots, b_n, we have \frac + \frac + \cdots + \frac \geq \frac. ( Nairi Sedrakyan (1997), Arthur Engel (1998), Titu Andreescu (2003))


Probabilistic statement

Similarly to the
Cauchy–Schwarz inequality The Cauchy–Schwarz inequality (also called Cauchy–Bunyakovsky–Schwarz inequality) is an upper bound on the absolute value of the inner product between two vectors in an inner product space in terms of the product of the vector norms. It is ...
, one can generalize Sedrakyan's inequality to
random variable A random variable (also called random quantity, aleatory variable, or stochastic variable) is a Mathematics, mathematical formalization of a quantity or object which depends on randomness, random events. The term 'random variable' in its mathema ...
s. In this formulation let X be a real random variable, and let Y be a positive random variable. ''X'' and ''Y'' need not be
independent Independent or Independents may refer to: Arts, entertainment, and media Artist groups * Independents (artist group), a group of modernist painters based in Pennsylvania, United States * Independentes (English: Independents), a Portuguese artist ...
, but we assume E /math> are both defined. Then \operatorname ^2/Y\ge \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities>X, /math> and E /math> are both defined. Then \operatorname ^2/Y\ge \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities>X, 2 / \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities>X, /math> and E /math> are both defined. Then \operatorname ^2/Y\ge \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities>X, /math> and E /math> are both defined. Then \operatorname ^2/Y\ge \operatorname X, 2 / \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities>X, 2 / \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities>X, 2 / \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities>X, 2 / \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities>X, /math> and E /math> are both defined. Then \operatorname ^2/Y\ge \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities>X, /math> and E /math> are both defined. Then \operatorname ^2/Y\ge \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities>X, 2 / \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities>X, /math> and E /math> are both defined. Then \operatorname ^2/Y\ge \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities>X, /math> and E /math> are both defined. Then \operatorname ^2/Y\ge \operatorname X, 2 / \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities>X, 2 / \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities>X, 2 / \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities>X, 2 / \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers ''a'', ''b'' and ''c'', :\frac + \frac + \frac \geq \frac, with equality only when a=b=c (i. e. in an equilateral triangle). There is no corre ...
. For positive real numbers a, b, c: \frac + \frac + \frac \geq \frac. Example 2.
International Mathematical Olympiad The International Mathematical Olympiad (IMO) is a mathematical olympiad for pre-university students, and is the oldest of the International Science Olympiads. It is widely regarded as the most prestigious mathematical competition in the wor ...
(IMO) 1995. For positive real numbers a,b,c , where abc=1 we have that \frac+\frac+\frac \geq \frac. Example 3. For positive real numbers a,b we have that 8(a^4+b^4) \geq (a+b)^4. Example 4. For positive real numbers a,b,c we have that \frac+\frac+\frac \geq \frac.


Proofs

Example 1. Proof: Use n = 3, \left(a_1, a_2, a_3\right) := (a, b, c), and \left(b_1, b_2, b_3\right) := (a(b + c), b(c + a), c(a + b)) to conclude: \frac + \frac + \frac \geq \frac = \frac = \frac + 1 \geq \frac (1) + 1 = \frac. \blacksquare Example 2. We have that \frac + \frac + \frac \geq \frac = \frac \geq \frac = \frac. Example 3. We have \frac + \frac \geq \frac so that a^4 + b^4 = \frac + \frac \geq \frac \geq \frac = \frac. Example 4. We have that \frac + \frac + \frac \geq \frac = \frac.


References

{{DEFAULTSORT:Sedrakyan's inequality Inequalities (mathematics) Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities