Nesbitt's Inequality

In mathematics, Nesbitt's 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, and was published at least 50 years earlier.

There is no corresponding upper bound as any of the 3 fractions in the inequality can be made arbitrarily large.

Proof

First proof: AM-HM inequality

By the AM- HM inequality on $(a+b),(b+c),(c+a)$,
:$\frac\geq\frac.$
Clearing denominators yields
:$((a+b)+(a+c)+(b+c))\left(\frac+\frac+\frac\right)\geq 9,$
from which we obtain
:$2\frac+2\frac+2\frac\geq9$
by expanding the product and collecting like denominators. This then simplifies directly to the final result.

Second proof: Rearrangement

Suppose $a \ge b \ge c$, we have that
:$\frac 1 \ge \frac 1 \ge \frac 1$
define
:$\vec x = (a, b, c)$
:$\vec y = \left(\frac 1 , \frac 1 , \frac 1 \right)$
The scalar product of the two sequences is maximum because of the rearrangement inequality if they are arranged the same way, call $\vec y_1$ and $\vec y_2$ the vector $\vec y$ shifted by one and by two, we have:
:$\vec x \cdot \vec y \ge \vec x \cdot \vec y_1$
:$\vec x \cdot \vec y \ge \vec x \cdot \vec y_2$

Addition yields our desired Nesbitt's inequality.

Third proof: Sum of Squares

The following identity is true for all $a,b,c:$

:$\frac+\frac+\frac = \frac + \frac \left(\frac+\frac+\frac\right)$
This clearly proves that the left side is no less than $\frac$ for positive a, b and c.

Note: every rational inequality can be demonstrated by transforming it to the appropriate sum-of-squares identity, see Hilbert's seventeenth problem.

Fourth proof: Cauchy–Schwarz

Invoking the Cauchy–Schwarz inequality on the vectors $\displaystyle\left\langle\sqrt,\sqrt,\sqrt\right\rangle,\left\langle\frac,\frac,\frac\right\rangle$ yields
:$((b+c)+(a+c)+(a+b))\left(\frac+\frac+\frac\right)\geq 9,$
which can be transformed into the final result as we did in the AM-HM proof.

Fifth proof: AM-GM

Let $x=a+b,y=b+c,z=c+a$. We then apply the AM-GM inequality to obtain the following
:$\frac+\frac+\frac\geq6.$
because \frac+\frac+\frac+\frac+\frac+\frac\geq6\sqrt 6.

Substituting out the $x,y,z$ in favor of $a,b,c$ yields
:$\frac+\frac+\frac\geq6$
:$\frac+\frac+\frac+3\geq6$
which then simplifies to the final result.

Sixth proof: Titu's lemma

Titu's lemma, a direct consequence of the Cauchy–Schwarz inequality, states that for any sequence of $n$ real numbers $(x_k)$ and any sequence of $n$ positive numbers $(a_k)$, $\displaystyle\sum_^n\frac\geq\frac$.

We use the lemma on $(x_k)=(1,1,1)$ and $(a_k)=(b+c,a+c,a+b)$. This gives,
:$\frac+\frac+\frac\geq\frac$
This results in,
:$\frac+\frac+\frac\geq\frac$ i.e.,
:$\frac+\frac+\frac\geq\frac-3=\frac$

Seventh proof: Using homogeneity

As the left side of the inequality is homogeneous, we may assume $a+b+c=1$. Now define $x=a+b$, $y=b+c$, and $z=c+a$. The desired inequality turns into $\frac+\frac+\frac\ge \frac$, or, equivalently, $\frac+\frac+\frac\ge 9/2$. This is clearly true by Titu's Lemma.

Eighth proof: Jensen inequality

Define $S=a+b+c$ and consider the function $f(x)=\frac$. This function can be shown to be convex in ,S/math> and, invoking Jensen inequality, we get
:$\displaystyle \frac\geq \frac.$
A straightforward computation yields
:$\frac+\frac+\frac\geq\frac.$

Ninth proof: Reduction to a two-variable inequality

By clearing denominators,
:$\frac+\frac+\frac\geq\frac\iff2(a^3+b^3+c^3)\geq ab^2+a^2b+ac^2+a^2c+bc^2+b^2c.$
It now suffices to prove that $x^3+y^3\geq xy^2+x^2y$ for $(x, y)\in\mathbb^2_+$, as summing this three times for $(x, y) = (a, b),\ (a, c),$ and $(b, c)$ completes the proof.

As $x^3+y^3\geq xy^2+x^2y\iff (x-y)(x^2-y^2)\geq 0$ we are done.

References

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* Ion Ionescu, Romanian Mathematical Gazette, Volume XXXII (September 15, 1926 - August 15, 1927), page 120
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External links

* Se
AoPS
for more proofs of this inequality.
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{{DEFAULTSORT:Nesbitt's Inequality
Inequalities