Nesbitt's Inequality

Revision as of 22:39, 3 December 2015 by Application (talk | contribs) (Proofs)

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

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

or equivalently

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

with equality when all the $a_i$ are equal.


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

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


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

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 applying AM-GM twice, we have

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

which yields the desired inequality.

By Expansion and AM-GM

We consider the equivalent inequality

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

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

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

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


The AM-HM inequality for three variables,

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

is equivalent to

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

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 $x+y+z \ge \frac{3}{2}$.

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

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

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 $(a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0$,

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


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

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

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

By Muirhead's and Cauchy

By Cauchy, \[\sum_{\text{cyc}}\frac{a^2}{ab + ac} \ge \frac{(a + b + c)^2}{2(ab + ac + bc)} = \frac{a^2 + b^2 + c^2 + 2(ab + ac + bc)}{2(ab + ac + bc)}\] \[= 1 + \frac{a^2 + b^2 + c^2}{2(ab + ac + bc)} \ge \frac{3}{2}\] since $a^2 + b^2 + c^2 \ge ab + ac + bc$ 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 \[\sum_{cyc} a^3+\sum_{sym} a^2b+\sum_{cyc} abc \ge \frac{3(a+b)(b+c)(c+a)}{2}\] This means \[[3,0,0]/2+[2,1,0]+[1,1,1]/2\ge\frac{3}{2}(a+b)(b+c)(a+c)\] So it follows that we must prove \[[3,0,0]+2[2,1,0]+[1,1,1]\ge3([2,1,0]+[1,1,1]/3)\] So it follows that we must prove $[3,0,0]\ge[2,1,0]$ which immediately follows from Muirheads.

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