Sedrakyan's Inequality
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The following inequality is known as Sedrakyan's inequality, Bergström's inequality, Engel's form or Titu's lemma, respectively, referring to the article ''About the applications of one useful inequality'' of
Nairi Sedrakyan Nairi Sedrakyan (born 1961 in Ninotsminda, USSR) is Erdős Award 2022 winner Armenian mathematician involved in national and international Olympiads, including American Mathematics Competitions (USA) and IMO, having been the president of the A ...
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 can be used as a mathematical proof technique and it has very useful new applications. In the book ''Algebraic Inequalities'' (Sedrakyan) are provided several generalizations of this inequality.


Statement of the inequality

For any reals 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 Nairi Sedrakyan (born 1961 in Ninotsminda, USSR) is Erdős Award 2022 winner Armenian mathematician involved in national and international Olympiads, including American Mathematics Competitions (USA) and IMO, having been the president of the A ...
(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 considered one of the most important and widely used inequalities in mathematics. The inequality for sums was published by . The corresponding inequality fo ...
, one can generalize Sedrakyan's inequality to
random variable A random variable (also called random quantity, aleatory variable, or stochastic variable) is a mathematical formalization of a quantity or object which depends on random events. It is a mapping or a function from possible outcomes (e.g., the po ...
. 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, but we assume E lt;_math>_are_both_defined. Then <math_display=block>.html" ;"title="X, ] and E /math> are both defined. Then \operatorname ^2/Y\ge \operatorname \ge_\operatorname 2___\operatorname <_math> <h1><br>.html" ;"title="X, ]^2 / \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 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 (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 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 (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 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 <_math> <h1><br>.html" ;"title="X, ]^2 / \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 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 (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 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 (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 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 (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 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 (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 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 <_math> <h1><br>.html" ;"title="X, ]^2 / \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 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 (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 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 (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 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 <_math> <h1><br>.html" ;"title="X, ]^2 / \operatorname \ge \operatorname 2 / \operatorname


Direct applications

Example 1.
Nesbitt's inequality In mathematics, Nesbitt's inequality (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 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 (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 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 (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 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 (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 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 (mathematics), inequality states that for positive real numbers ''a'', ''b'' and ''c'', :\frac+\frac+\frac\geq\frac. It is an elementary special case (N = 3) of the difficult and much studied Shapiro inequality, ...
. 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. The first IMO was held in Romania in 1959. It has since been held annually, except i ...
(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 Linear algebra Operator theory Articles containing proofs Mathematical analysis Probabilistic inequalities