Difference between revisions of "Nesbitt's Inequality"

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'''Nesbitt's [[Inequality]]''' is a theorem which, although rarely cited, has many instructive proofs.  It states that for positive <math> \displaystyle a, b, c </math>,
+
'''Nesbitt's [[Inequality]]''' is a theorem which, although rarely cited, has many instructive proofs.  It states that for positive <math>a, b, c </math>,
 
<center>
 
<center>
 
<math>
 
<math>
Line 12: Line 12:
 
<center>
 
<center>
 
<math>
 
<math>
\sum_{i=1}^{n}\frac{a_i}{s-a_i} \ge \frac{n}{n-1}
+
\sum_{i=1}^{n}\frac{a_i}{s-a_i} \geq \frac{n}{n-1}
 
</math>,
 
</math>,
 
</center>
 
</center>
Line 18: Line 18:
 
<center>
 
<center>
 
<math>
 
<math>
\sum_{i=1}^{n}\frac{s}{s-a_i} \ge \frac{n^2}{n-1}
+
\sum_{i=1}^{n}\frac{s}{s-a_i} \geq \frac{n^2}{n-1}
 
</math>,
 
</math>,
 
</center>
 
</center>
with equality when all the <math> \displaystyle a_i </math> are equal.
+
with equality when all the <math>a_i </math> are equal.
  
 
== Proofs ==
 
== Proofs ==
 +
 +
=== By Smoothing ===
 +
 +
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.
 +
 +
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 ===
  
Note that <math> \displaystyle 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]],
+
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]],
 
<center>
 
<center>
 
<math>
 
<math>
Line 33: Line 39:
 
</math>.
 
</math>.
 
</center>
 
</center>
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> \displaystyle a=b </math>, so by symmetry, all the variables must be equal.
+
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.
  
 
=== By Cauchy ===
 
=== By Cauchy ===
Line 40: Line 46:
 
<center>
 
<center>
 
<math>
 
<math>
[(b+c) + (c+a) + (a+b)]\left( \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \right) \ge 9
+
[(b+c) + (c+a) + (a+b)]\left( \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \right) \geq 9
 
</math>,
 
</math>,
 
</center>
 
</center>
Line 46: Line 52:
 
<center>
 
<center>
 
<math>
 
<math>
2\left( \frac{a+b+c}{b+c} + \frac{a+b+c}{c+a} + \frac{a+b+c}{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) \geq 9
 
</math>,
 
</math>,
 
</center>
 
</center>
as desired.  Equality occurs when <math> \displaystyle (b+c)^2 = (c+a)^2 = (a+b)^2 </math>, i.e., when <math> \displaystyle a=b=c </math>.
+
as desired.  Equality occurs when <math>(b+c)^2 = (c+a)^2 = (a+b)^2 </math>, i.e., when <math>a=b=c </math>.
  
 
We also present three closely related variations of this proof, which illustrate how [[AM-HM]] is related to [[AM-GM]] and Cauchy.
 
We also present three closely related variations of this proof, which illustrate how [[AM-HM]] is related to [[AM-GM]] and Cauchy.
Line 58: Line 64:
 
<center>
 
<center>
 
<math>
 
<math>
[(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
+
[(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
 
</math>,
 
</math>,
 
</center>
 
</center>
Line 68: Line 74:
 
<center>
 
<center>
 
<math>
 
<math>
[(b+c) + (c+a) + (a+c)]\left( \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \right) \ge 9
+
[(b+c) + (c+a) + (a+c)]\left( \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \right) \geq 9
 
</math>.
 
</math>.
 
</center>
 
</center>
Setting <math> \displaystyle x = b+c, y= c+a, z= a+b </math>, we expand the left side to obtain
+
Setting <math>x = b+c, y= c+a, z= a+b </math>, we expand the left side to obtain
 
<center>
 
<center>
 
<math>
 
<math>
3 + \frac{x}{y} + \frac{y}{x} + \frac{y}{z} + \frac{z}{y} + \frac{z}{x} + \frac{x}{z} \ge 9
+
3 + \frac{x}{y} + \frac{y}{x} + \frac{y}{z} + \frac{z}{y} + \frac{z}{x} + \frac{x}{z} \geq 9
 
</math>,
 
</math>,
 
</center>
 
</center>
which follows from <math> \frac{x}{y} + \frac{y}{x} \ge 2 </math>, etc., by [[AM-GM]], with equality when <math> \displaystyle x=y=z </math>.
+
which follows from <math> \frac{x}{y} + \frac{y}{x} \geq 2 </math>, etc., by [[AM-GM]], with equality when <math>x=y=z </math>.
  
 
==== By AM-HM ====
 
==== By AM-HM ====
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<center>
 
<center>
 
<math>
 
<math>
\frac{x+y+z}{3} \ge \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}
+
\frac{x+y+z}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}}
 
</math>,
 
</math>,
 
</center>
 
</center>
Line 90: Line 96:
 
<center>
 
<center>
 
<math>
 
<math>
(x+y+z) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \ge 9
+
(x+y+z) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \geq 9
 
</math>.
 
</math>.
 
</center>
 
</center>
Setting <math> \displaystyle x=b+c, y=c+a, z=a+b </math> yields the desired inequality.
+
Setting <math>x=b+c, y=c+a, z=a+b </math> yields the desired inequality.
  
 
=== By Substitution ===
 
=== By Substitution ===
  
The numbers <math> \displaystyle x = \frac{a}{b+c}, y = \frac{b}{c+a}, z = \frac{c}{a+b} </math> satisfy the condition <math> \displaystyle xy + yz + zx + 2xyz = 1 </math>.  Thus it is sufficient to prove that if any numbers <math> \displaystyle x,y,z </math> satisfy <math> \displaystyle xy + yz + zx + 2xyz = 1 </math>, then <math> x+y+z \ge \frac{3}{2} </math>.
+
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 \geq \frac{3}{2} </math>.
  
Suppose, on the contrary, that <math> x+y+z < \frac{3}{2} </math>.  We then have <math> \displaystyle xy + yz + zx \le \left( \frac{x+y+z}{3} \right)^2 < \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> \displaystyle xy + yz + zx + 2xyz < 1 </math>, a contradiction.
+
Suppose, on the contrary, that <math> x+y+z < \frac{3}{2} </math>.  We then have <math>xy + yz + zx \leq \frac{(x+y+z)^2}{3} < \frac{3}{4} </math>, and <math> 2xyz \leq 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.
  
 
=== By Normalization and AM-HM ===
 
=== By Normalization and AM-HM ===
  
We may normalize so that <math> \displaystyle a+b+c = 1 </math>.  It is then sufficient to prove
+
We may normalize so that <math>a+b+c = 1 </math>.  It is then sufficient to prove
 
<center>
 
<center>
 
<math>
 
<math>
\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \ge \frac{9}{2}
+
\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \geq \frac{9}{2}
 
</math>,
 
</math>,
 
</center>
 
</center>
Line 113: Line 119:
 
=== By Weighted AM-HM ===
 
=== By Weighted AM-HM ===
  
We may normalize so that <math> \displaystyle a+b+c =1 </math>.
+
We may normalize so that <math>a+b+c =1 </math>.
  
We first note that by the [[rearrangement inequality]],
+
We first note that by the [[rearrangement inequality]] or the fact that <math>(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0</math>,
 
<center>
 
<center>
 
<math>
 
<math>
3 (ab + bc + ca) \le a^2 + b^2 + c^2 + 2(ab + bc + ca)  
+
3 (ab + bc + ca) \leq a^2 + b^2 + c^2 + 2(ab + bc + ca)  
 
</math>,
 
</math>,
 
</center>
 
</center>
Line 124: Line 130:
 
<center>
 
<center>
 
<math>
 
<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}
+
\frac{1}{a(b+c) + b(c+a) + c(a+b)} \geq \frac{1}{\frac{2}{3}(a+b+c)^2} = \frac{3}{2}
 
</math>.
 
</math>.
 
</center>
 
</center>
  
Since <math> \displaystyle a+b+c = 1 </math>, weighted AM-HM gives us
+
Since <math>a+b+c = 1 </math>, weighted AM-HM gives us
 
<center>
 
<center>
 
<math>
 
<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}
+
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}
 
</math>.
 
</math>.
 
</center>
 
</center>
 +
 +
===By Muirhead's and Cauchy===
 +
 +
By Cauchy, <cmath>\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)}</cmath> <cmath> = 1 + \frac{a^2 + b^2 + c^2}{2(ab + ac + bc)} \geq \frac{3}{2}</cmath> since <math>a^2 + b^2 + c^2 \geq ab + ac + bc</math> by Muirhead as <math>[1, 1, 0]\prec [2, 0, 0]</math>
 +
 +
===Another Interesting Method===
 +
 +
Let <cmath>S=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}</cmath>
 +
And <cmath>M=\frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}</cmath>
 +
And <cmath>N=\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}</cmath>
 +
Now, we get <cmath>M+N=3</cmath>
 +
Also by AM-GM; <cmath>M+S\geq 3</cmath> and <cmath>N+S\geq 3</cmath>
 +
<cmath>\implies M+N+2S\geq 6</cmath>
 +
<cmath>\implies 2S\geq 3</cmath>
 +
<cmath>\implies S\geq \frac{3}{2}</cmath>
 +
 +
===By Muirhead's and expansion===
 +
 +
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 \geq \frac{3(a+b)(b+c)(c+a)}{2}</cmath> This means <cmath>[3,0,0]/2+[2,1,0]+[1,1,1]/2\geq \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]\geq 3([2,1,0]+[1,1,1]/3)</cmath> So it follows that we must prove <math>[3,0,0]\geq [2,1,0]</math> which immediately follows from Muirhead's.
 +
[[Category:Inequality]]
 +
[[Category:Theorems]]

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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