Difference between revisions of "Nesbitt's Inequality"

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== Proofs ==
 
== Proofs ==
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=== By Smoothing ===
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We may normalize so that <math>a+b+c=1</math>. Then we wish to show<cmath>\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq\frac32.</cmath>The equality case is when <math>a=b=c</math> so we can use smoothing. We wish to show that if <math>a\neq b</math> then<cmath>\frac{a}{1-a}+\frac{b}{1-b}>2\cdot\frac{\frac{a+b}2}{1-\frac{a+b}2}.</cmath>If this is true, then if the minimum occurs at a point where <math>a\neq b</math> then we can replace <math>a</math> and <math>b</math> with their average and then the expression decreases, so that cannot be the minimum. Thus the minimum occurs where they are all equal, and we would be done.
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Expanding both sides of the inequality and clearing denominators we get <cmath>a^2+b^2>2ab,</cmath>which is true because <math>a\neq b</math>.
  
 
=== By Rearrangement ===
 
=== By Rearrangement ===

Latest revision as of 16:41, 3 March 2020

Nesbitt's Inequality is a theorem which, although rarely cited, has many instructive proofs. It states that for positive $a, b, c$,

$\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \ge \frac{3}{2}$,

with equality when all the variables are equal.

All of the proofs below generalize to proof the following more general inequality.

If $a_1, \ldots a_n$ are positive and $\sum_{i=1}^{n}a_i = s$, then

$\sum_{i=1}^{n}\frac{a_i}{s-a_i} \geq \frac{n}{n-1}$,

or equivalently

$\sum_{i=1}^{n}\frac{s}{s-a_i} \geq \frac{n^2}{n-1}$,

with equality when all the $a_i$ are equal.

Proofs

By Smoothing

We may normalize so that $a+b+c=1$. Then we wish to show\[\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\geq\frac32.\]The equality case is when $a=b=c$ so we can use smoothing. We wish to show that if $a\neq b$ then\[\frac{a}{1-a}+\frac{b}{1-b}>2\cdot\frac{\frac{a+b}2}{1-\frac{a+b}2}.\]If this is true, then if the minimum occurs at a point where $a\neq b$ then we can replace $a$ and $b$ with their average and then the expression decreases, so that cannot be the minimum. Thus the minimum occurs where they are all equal, and we would be done.

Expanding both sides of the inequality and clearing denominators we get \[a^2+b^2>2ab,\]which is true because $a\neq b$.

By Rearrangement

Note that $a,b,c$ and $\frac{1}{b+c} = \frac{1}{a+b+c -a}$, $\frac{1}{c+a} = \frac{1}{a+b+c -b}$, $\frac{1}{a+b} = \frac{1}{a+b+c -c}$ are sorted in the same order. Then by the rearrangement inequality,

$2 \left( \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \right) \ge \frac{b}{b+c} + \frac{c}{b+c} + \frac{c}{c+a} + \frac{a}{c+a} + \frac{a}{a+b} + \frac{b}{a+b} = 3$.

For equality to occur, since we changed ${} a \cdot \frac{1}{b+c} + b \cdot \frac{1}{c+a}$ to $b \cdot \frac{1}{b+c} + a \cdot \frac{1}{c+a}$, we must have $a=b$, so by symmetry, all the variables must be equal.

By Cauchy

By the Cauchy-Schwarz Inequality, we have

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

or

$2\left( \frac{a+b+c}{b+c} + \frac{a+b+c}{c+a} + \frac{a+b+c}{a+b} \right) \geq 9$,

as desired. Equality occurs when $(b+c)^2 = (c+a)^2 = (a+b)^2$, i.e., when $a=b=c$.

We also present three closely related variations of this proof, which illustrate how AM-HM is related to AM-GM and Cauchy.

By AM-GM

By applying AM-GM twice, we have

$[(b+c) + (c+a) + (a+b)] \left( \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \right) \geq 3 [(b+c)(c+a)(a+b)]^{\frac{1}{3}} \cdot \left(\frac{1}{(b+c)(c+a)(a+b)}\right)^{\frac{1}{3}} = 9$,

which yields the desired inequality.

By Expansion and AM-GM

We consider the equivalent inequality

$[(b+c) + (c+a) + (a+c)]\left( \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \right) \geq 9$.

Setting $x = b+c, y= c+a, z= a+b$, we expand the left side to obtain

$3 + \frac{x}{y} + \frac{y}{x} + \frac{y}{z} + \frac{z}{y} + \frac{z}{x} + \frac{x}{z} \geq 9$,

which follows from $\frac{x}{y} + \frac{y}{x} \geq 2$, etc., by AM-GM, with equality when $x=y=z$.

By AM-HM

The AM-HM inequality for three variables,

$\frac{x+y+z}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}$,

is equivalent to

$(x+y+z) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \geq 9$.

Setting $x=b+c, y=c+a, z=a+b$ yields the desired inequality.

By Substitution

The numbers $x = \frac{a}{b+c}, y = \frac{b}{c+a}, z = \frac{c}{a+b}$ satisfy the condition $xy + yz + zx + 2xyz = 1$. Thus it is sufficient to prove that if any numbers $x,y,z$ satisfy $xy + yz + zx + 2xyz = 1$, then $x+y+z \geq \frac{3}{2}$.

Suppose, on the contrary, that $x+y+z < \frac{3}{2}$. We then have $xy + yz + zx \leq \frac{(x+y+z)^2}{3} < \frac{3}{4}$, and $2xyz \leq 2 \left( \frac{x+y+z}{3} \right)^3 < \frac{1}{4}$. Adding these inequalities yields $xy + yz + zx + 2xyz < 1$, a contradiction.

By Normalization and AM-HM

We may normalize so that $a+b+c = 1$. It is then sufficient to prove

$\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \geq \frac{9}{2}$,

which follows from AM-HM.

By Weighted AM-HM

We may normalize so that $a+b+c =1$.

We first note that by the rearrangement inequality or the fact that $(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$,

$3 (ab + bc + ca) \leq a^2 + b^2 + c^2 + 2(ab + bc + ca)$,

so

$\frac{1}{a(b+c) + b(c+a) + c(a+b)} \geq \frac{1}{\frac{2}{3}(a+b+c)^2} = \frac{3}{2}$.

Since $a+b+c = 1$, weighted AM-HM gives us

$a\cdot \frac{1}{b+c} + b \cdot \frac{1}{c+a} + c \cdot \frac{1}{a+b} \geq \frac{1}{a(b+c) + b(c+a) + c(a+b)} \geq \frac{3}{2}$.

By Muirhead's and Cauchy

By Cauchy, \[\sum_{\text{cyc}}\frac{a^2}{ab + ac} \geq \frac{(a + b + c)^2}{2(ab + ac + bc)} = \frac{a^2 + b^2 + c^2 + 2(ab + ac + bc)}{2(ab + ac + bc)}\] \[= 1 + \frac{a^2 + b^2 + c^2}{2(ab + ac + bc)} \geq \frac{3}{2}\] since $a^2 + b^2 + c^2 \geq ab + ac + bc$ by Muirhead as $[1, 1, 0]\prec [2, 0, 0]$

Another Interesting Method

Let \[S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\] And \[M=\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}\] And \[N=\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}\] Now, we get \[M+N=3\] Also by AM-GM; \[M+S\geq 3\] and \[N+S\geq 3\] \[\implies M+N+2S\geq 6\] \[\implies 2S\geq 3\] \[\implies S\geq \frac{3}{2}\]

By Muirhead's and expansion

Let $[x,y,z]=\sum_{sym} a^xb^yc^z$. Expanding out we get that our inequality is equivalent to \[\sum_{cyc} a^3+\sum_{sym} a^2b+\sum_{cyc} abc \geq \frac{3(a+b)(b+c)(c+a)}{2}\] This means \[[3,0,0]/2+[2,1,0]+[1,1,1]/2\geq \frac{3}{2}(a+b)(b+c)(a+c)\] So it follows that we must prove \[[3,0,0]+2[2,1,0]+[1,1,1]\geq 3([2,1,0]+[1,1,1]/3)\] So it follows that we must prove $[3,0,0]\geq [2,1,0]$ which immediately follows from Muirhead's.

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