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Nesbitt's Inequality: Difference between revisions

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added another proof (original?)
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with equality when all the variables are equal.
with equality when all the variables are equal.


Most of the proofs below generalize to proof the following stronger inequality.
All of the proofs below generalize to proof the following stronger inequality.


If <math> a_1, \ldots a_n </math> are positive and <math> \sum_{i=1}^{n}a_i = s </math>, then  
If <math> a_1, \ldots a_n </math> are positive and <math> \sum_{i=1}^{n}a_i = s </math>, then  
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</center>
</center>
which follows from [[AM-HM]].
which follows from [[AM-HM]].
=== By Weighted AM-HM ===
We may normalize so that <math> \displaystyle a+b+c =1 </math>.
We first note that by the [[rearrangement inequality]],
<center>
<math>
3 (ab + bc + ca) \le a^2 + b^2 + c^2 + 2(ab + bc + ca)
</math>,
</center>
so
<center>
<math>
\frac{1}{a(b+c) + b(c+a) + c(a+b)} \ge \frac{1}{\frac{2}{3}(a+b+c)^2} = \frac{3}{2}
</math>.
</center>
Since <math> \displaystyle a+b+c = 1 </math>, weighted AM-HM gives us
<center>
<math>
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}
</math>.
</center>

Revision as of 15:31, 29 April 2007

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

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

with equality when all the variables are equal.

All of the proofs below generalize to proof the following stronger 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 $\displaystyle a_i$ are equal.

Proofs

By Rearrangement

Note that $\displaystyle a,b,c$ 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 $\displaystyle a=b$, so by symmetry, all the variables must be equal.

By Cauchy

By the Cauchy-Schwarz Inequality, we have

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

or

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

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

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

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

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 $\displaystyle x = b+c, y= c+a, z= a+b$, 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 $\displaystyle x=y=z$.

By AM-HM

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

By Substitution

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

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

By Normalization and AM-HM

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

We first note that by the rearrangement inequality,

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

so

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

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

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

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