Nesbitt's inequality

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In mathematics, Nesbitt's inequality, named after Alfred Nesbitt, states that for positive real numbers a, b and c,

[math]\displaystyle{ \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}. }[/math]

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 [math]\displaystyle{ (a+b),(b+c),(c+a) }[/math],

[math]\displaystyle{ \frac{(a+b)+(a+c)+(b+c)}{3}\geq\frac{3}{\displaystyle\frac{1}{a+b}+\frac{1}{a+c}+ \frac{1}{b+c}}. }[/math]

Clearing denominators yields

[math]\displaystyle{ ((a+b)+(a+c)+(b+c))\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)\geq 9, }[/math]

from which we obtain

[math]\displaystyle{ 2\frac{a+b+c}{b+c}+2\frac{a+b+c}{a+c}+2\frac{a+b+c}{a+b}\geq9 }[/math]

by expanding the product and collecting like denominators. This then simplifies directly to the final result.

Second proof: Rearrangement

Suppose [math]\displaystyle{ a \ge b \ge c }[/math], we have that

[math]\displaystyle{ \frac 1 {b+c} \ge \frac 1 {a+c} \ge \frac 1 {a+b} }[/math]

define

[math]\displaystyle{ \vec x = (a, b, c) }[/math]
[math]\displaystyle{ \vec y = \left(\frac 1 {b+c} , \frac 1 {a+c} , \frac 1 {a+b}\right) }[/math]

The scalar product of the two sequences is maximum because of the rearrangement inequality if they are arranged the same way, call [math]\displaystyle{ \vec y_1 }[/math] and [math]\displaystyle{ \vec y_2 }[/math] the vector [math]\displaystyle{ \vec y }[/math] shifted by one and by two, we have:

[math]\displaystyle{ \vec x \cdot \vec y \ge \vec x \cdot \vec y_1 }[/math]
[math]\displaystyle{ \vec x \cdot \vec y \ge \vec x \cdot \vec y_2 }[/math]

Addition yields our desired Nesbitt's inequality.

Third proof: Sum of Squares

The following identity is true for all [math]\displaystyle{ a,b,c: }[/math]

[math]\displaystyle{ \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} = \frac{3}{2} + \frac{1}{2} \left(\frac{(a-b)^2}{(a+c)(b+c)}+\frac{(a-c)^2}{(a+b)(b+c)}+\frac{(b-c)^2}{(a+b)(a+c)}\right) }[/math]

This clearly proves that the left side is no less than [math]\displaystyle{ \frac{3}{2} }[/math] 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 [math]\displaystyle{ \displaystyle\left\langle\sqrt{a+b},\sqrt{b+c},\sqrt{c+a}\right\rangle,\left\langle\frac{1}{\sqrt{a+b}},\frac{1}{\sqrt{b+c}},\frac{1}{\sqrt{c+a}}\right\rangle }[/math] yields

[math]\displaystyle{ ((b+c)+(a+c)+(a+b))\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\geq 9, }[/math]

which can be transformed into the final result as we did in the AM-HM proof.

Fifth proof: AM-GM

Let [math]\displaystyle{ x=a+b,y=b+c,z=c+a }[/math]. We then apply the AM-GM inequality to obtain the following

[math]\displaystyle{ \frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}\geq6. }[/math]

because [math]\displaystyle{ \frac{x}{y}+\frac{z}{y}+\frac{y}{x}+\frac{z}{x}+\frac{x}{z}+\frac{y}{z}\geq6\sqrt[6]{\frac{x}{y}\cdot\frac{z}{y}\cdot\frac{y}{x}\cdot\frac{z}{x}\cdot\frac{x}{z}\cdot \frac{y}{z}}=6. }[/math]

Substituting out the [math]\displaystyle{ x,y,z }[/math] in favor of [math]\displaystyle{ a,b,c }[/math] yields

[math]\displaystyle{ \frac{2a+b+c}{b+c}+\frac{a+b+2c}{a+b}+\frac{a+2b+c}{c+a}\geq6 }[/math]
[math]\displaystyle{ \frac{2a}{b+c}+\frac{2c}{a+b}+\frac{2b}{a+c}+3\geq6 }[/math]

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 [math]\displaystyle{ n }[/math] real numbers [math]\displaystyle{ (x_k) }[/math] and any sequence of [math]\displaystyle{ n }[/math] positive numbers [math]\displaystyle{ (a_k) }[/math], [math]\displaystyle{ \displaystyle\sum_{k=1}^n\frac{x_k^2}{a_k}\geq\frac{(\sum_{k=1}^n x_k)^2}{\sum_{k=1}^n a_k} }[/math].

We use the lemma on [math]\displaystyle{ (x_k)=(1,1,1) }[/math] and [math]\displaystyle{ (a_k)=(b+c,a+c,a+b) }[/math]. This gives,

[math]\displaystyle{ \frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\geq\frac{3^2}{2(a+b+c)} }[/math]

This results in,

[math]\displaystyle{ \frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b}\geq\frac{9}{2} }[/math] i.e.,
[math]\displaystyle{ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{9}{2}-3=\frac{3}{2} }[/math]

Seventh proof: Using homogeneity

As the left side of the inequality is homogeneous, we may assume [math]\displaystyle{ a+b+c=1 }[/math]. Now define [math]\displaystyle{ x=a+b }[/math], [math]\displaystyle{ y=b+c }[/math], and [math]\displaystyle{ z=c+a }[/math]. The desired inequality turns into [math]\displaystyle{ \frac{1-x}{x}+\frac{1-y}{y}+\frac{1-z}{z}\ge \frac{3}{2} }[/math], or, equivalently, [math]\displaystyle{ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge 9/2 }[/math]. This is clearly true by Titu's Lemma.

Eighth proof: Jensen inequality

Define [math]\displaystyle{ S=a+b+c }[/math] and consider the function [math]\displaystyle{ f(x)=\frac{x}{S-x} }[/math]. This function can be shown to be convex in [math]\displaystyle{ [0,S] }[/math] and, invoking Jensen inequality, we get

[math]\displaystyle{ \displaystyle \frac{\frac{a}{S-a}+\frac{b}{S-b}+\frac{c}{S-c}}{3}\geq \frac{S/3}{S-S/3}. }[/math]

A straightforward computation yields

[math]\displaystyle{ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}. }[/math]

Ninth proof: Reduction to a two-variable inequality

By clearing denominators,

[math]\displaystyle{ \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq\frac{3}{2}\iff2(a^3+b^3+c^3)\geq ab^2+a^2b+ac^2+a^2c+bc^2+b^2c. }[/math]

It now suffices to prove that [math]\displaystyle{ x^3+y^3\geq xy^2+x^2y }[/math] for [math]\displaystyle{ (x, y)\in\mathbb{R}^2_+ }[/math], as summing this three times for [math]\displaystyle{ (x, y) = (a, b),\ (a, c), }[/math] and [math]\displaystyle{ (b, c) }[/math] completes the proof.

As [math]\displaystyle{ x^3+y^3\geq xy^2+x^2y\iff (x-y)(x^2-y^2)\geq 0 }[/math] we are done.

References

External links



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