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Nesbitt's Inequality
2015-12-04T03:39:17Z
<p>Application: /* Proofs */</p>
<hr />
<div>'''Nesbitt's [[Inequality]]''' is a theorem which, although rarely cited, has many instructive proofs. It states that for positive <math>a, b, c </math>,<br />
<center><br />
<math><br />
\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \ge \frac{3}{2}<br />
</math>,<br />
</center><br />
with equality when all the variables are equal.<br />
<br />
All of the proofs below generalize to proof the following more general inequality.<br />
<br />
If <math> a_1, \ldots a_n </math> are positive and <math> \sum_{i=1}^{n}a_i = s </math>, then <br />
<center><br />
<math><br />
\sum_{i=1}^{n}\frac{a_i}{s-a_i} \ge \frac{n}{n-1}<br />
</math>,<br />
</center><br />
or equivalently<br />
<center><br />
<math><br />
\sum_{i=1}^{n}\frac{s}{s-a_i} \ge \frac{n^2}{n-1}<br />
</math>,<br />
</center><br />
with equality when all the <math>a_i </math> are equal.<br />
<br />
== Proofs ==<br />
<br />
=== By Rearrangement ===<br />
<br />
Note that <math>a,b,c </math> and <math> \frac{1}{b+c} = \frac{1}{a+b+c -a}</math>, <math> \frac{1}{c+a} = \frac{1}{a+b+c -b} </math>, <math> \frac{1}{a+b} = \frac{1}{a+b+c -c} </math> are sorted in the same order. Then by the [[rearrangement inequality]],<br />
<center><br />
<math><br />
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<br />
</math>.<br />
</center><br />
For equality to occur, since we changed <math>{} a \cdot \frac{1}{b+c} + b \cdot \frac{1}{c+a} </math> to <math> b \cdot \frac{1}{b+c} + a \cdot \frac{1}{c+a} </math>, we must have <math>a=b </math>, so by symmetry, all the variables must be equal.<br />
<br />
=== By Cauchy ===<br />
<br />
By the [[Cauchy-Schwarz Inequality]], we have<br />
<center><br />
<math><br />
[(b+c) + (c+a) + (a+b)]\left( \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \right) \ge 9<br />
</math>,<br />
</center><br />
or<br />
<center><br />
<math><br />
2\left( \frac{a+b+c}{b+c} + \frac{a+b+c}{c+a} + \frac{a+b+c}{a+b} \right) \ge 9<br />
</math>,<br />
</center><br />
as desired. Equality occurs when <math>(b+c)^2 = (c+a)^2 = (a+b)^2 </math>, i.e., when <math>a=b=c </math>.<br />
<br />
We also present three closely related variations of this proof, which illustrate how [[AM-HM]] is related to [[AM-GM]] and Cauchy.<br />
<br />
==== By AM-GM ====<br />
<br />
By applying [[AM-GM]] twice, we have<br />
<center><br />
<math><br />
[(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<br />
</math>,<br />
</center><br />
which yields the desired inequality.<br />
<br />
==== By Expansion and AM-GM ====<br />
<br />
We consider the equivalent inequality<br />
<center><br />
<math><br />
[(b+c) + (c+a) + (a+c)]\left( \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \right) \ge 9<br />
</math>.<br />
</center><br />
Setting <math>x = b+c, y= c+a, z= a+b </math>, we expand the left side to obtain<br />
<center><br />
<math><br />
3 + \frac{x}{y} + \frac{y}{x} + \frac{y}{z} + \frac{z}{y} + \frac{z}{x} + \frac{x}{z} \ge 9<br />
</math>,<br />
</center><br />
which follows from <math> \frac{x}{y} + \frac{y}{x} \ge 2 </math>, etc., by [[AM-GM]], with equality when <math>x=y=z </math>.<br />
<br />
==== By AM-HM ====<br />
<br />
The [[AM-HM]] inequality for three variables,<br />
<center><br />
<math><br />
\frac{x+y+z}{3} \ge \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}<br />
</math>,<br />
</center><br />
is equivalent to<br />
<center><br />
<math><br />
(x+y+z) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \ge 9<br />
</math>.<br />
</center><br />
Setting <math>x=b+c, y=c+a, z=a+b </math> yields the desired inequality.<br />
<br />
=== By Substitution ===<br />
<br />
The numbers <math>x = \frac{a}{b+c}, y = \frac{b}{c+a}, z = \frac{c}{a+b} </math> satisfy the condition <math>xy + yz + zx + 2xyz = 1 </math>. Thus it is sufficient to prove that if any numbers <math>x,y,z </math> satisfy <math>xy + yz + zx + 2xyz = 1 </math>, then <math> x+y+z \ge \frac{3}{2} </math>.<br />
<br />
Suppose, on the contrary, that <math> x+y+z < \frac{3}{2} </math>. We then have <math>xy + yz + zx \le \frac{(x+y+z)^2}{3} < \frac{3}{4} </math>, and <math> 2xyz \le 2 \left( \frac{x+y+z}{3} \right)^3 < \frac{1}{4} </math>. Adding these inequalities yields <math>xy + yz + zx + 2xyz < 1 </math>, a contradiction.<br />
<br />
=== By Normalization and AM-HM ===<br />
<br />
We may normalize so that <math>a+b+c = 1 </math>. It is then sufficient to prove<br />
<center><br />
<math><br />
\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \ge \frac{9}{2}<br />
</math>,<br />
</center><br />
which follows from [[AM-HM]].<br />
<br />
=== By Weighted AM-HM ===<br />
<br />
We may normalize so that <math>a+b+c =1 </math>.<br />
<br />
We first note that by the [[rearrangement inequality]] or the fact that <math>(a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0</math>,<br />
<center><br />
<math><br />
3 (ab + bc + ca) \le a^2 + b^2 + c^2 + 2(ab + bc + ca) <br />
</math>,<br />
</center><br />
so<br />
<center><br />
<math><br />
\frac{1}{a(b+c) + b(c+a) + c(a+b)} \ge \frac{1}{\frac{2}{3}(a+b+c)^2} = \frac{3}{2}<br />
</math>.<br />
</center><br />
<br />
Since <math>a+b+c = 1 </math>, weighted AM-HM gives us<br />
<center><br />
<math><br />
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}<br />
</math>.<br />
</center><br />
<br />
===By Muirhead's and Cauchy===<br />
<br />
By Cauchy, <cmath>\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)}</cmath> <cmath> = 1 + \frac{a^2 + b^2 + c^2}{2(ab + ac + bc)} \ge \frac{3}{2}</cmath> since <math>a^2 + b^2 + c^2 \ge ab + ac + bc</math> by Muirhead as <math>[1, 1, 0]\prec [2, 0, 0]</math><br />
<br />
===Another Interesting Method===<br />
<br />
Let <cmath>S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}</cmath><br />
And <cmath>M=\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}</cmath><br />
And <cmath>N=\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}</cmath><br />
Now, we get <cmath>M+N=3</cmath><br />
Also by AM-GM; <cmath>M+S\geq 3</cmath> and <cmath>N+S\geq 3</cmath><br />
<cmath>\implies M+N+2S\geq 6</cmath><br />
<cmath>\implies 2S\geq 3</cmath><br />
<cmath>\implies S\geq \frac{3}{2}</cmath><br />
<br />
===By Muirhead's and expansion===<br />
<br />
Let <math>[x,y,z]=\sum_{sym} a^xb^yc^z</math>. Expanding out we get that our inequality is equivalent to <cmath>\sum_{cyc} a^3+\sum_{sym} a^2b+\sum_{cyc} abc \ge \frac{3(a+b)(b+c)(c+a)}{2}</cmath> This means <cmath>[3,0,0]/2+[2,1,0]+[1,1,1]/2\ge\frac{3}{2}(a+b)(b+c)(a+c)</cmath> So it follows that we must prove <cmath>[3,0,0]+2[2,1,0]+[1,1,1]\ge3([2,1,0]+[1,1,1]/3)</cmath> So it follows that we must prove <math>[3,0,0]\ge[2,1,0]</math> which immediately follows from Muirheads.<br />
[[Category:Inequality]]<br />
[[Category:Theorems]]</div>
Application