Difference between revisions of "Nesbitt's Inequality"
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with equality when all the variables are equal. | with equality when all the variables are equal. | ||
− | All of the proofs below generalize to | + | All of the proofs below generalize to prove the following more general 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 | ||
<center> | <center> | ||
<math> | <math> | ||
− | \sum_{i=1}^{n}\frac{a_i}{s-a_i} \ | + | \sum_{i=1}^{n}\frac{a_i}{s-a_i} \geq \frac{n}{n-1} |
</math>, | </math>, | ||
</center> | </center> | ||
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<center> | <center> | ||
<math> | <math> | ||
− | \sum_{i=1}^{n}\frac{s}{s-a_i} \ | + | \sum_{i=1}^{n}\frac{s}{s-a_i} \geq \frac{n^2}{n-1} |
</math>, | </math>, | ||
</center> | </center> | ||
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== 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 === | ||
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<center> | <center> | ||
<math> | <math> | ||
− | [(b+c) + (c+a) + (a+b)]\left( \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \right) \ | + | [(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> | ||
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<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) \ | + | 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> | ||
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<center> | <center> | ||
<math> | <math> | ||
− | [(b+c) + (c+a) + (a+b)] \left( \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \right) \ | + | [(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> | ||
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<center> | <center> | ||
<math> | <math> | ||
− | [(b+c) + (c+a) + (a+c)]\left( \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \right) \ | + | [(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> | ||
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<center> | <center> | ||
<math> | <math> | ||
− | 3 + \frac{x}{y} + \frac{y}{x} + \frac{y}{z} + \frac{z}{y} + \frac{z}{x} + \frac{x}{z} \ | + | 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} \ | + | 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} \ | + | \frac{x+y+z}{3} \geq \frac{3}{\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} |
</math>, | </math>, | ||
</center> | </center> | ||
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<center> | <center> | ||
<math> | <math> | ||
− | (x+y+z) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \ | + | (x+y+z) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \geq 9 |
</math>. | </math>. | ||
</center> | </center> | ||
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=== By Substitution === | === By Substitution === | ||
− | 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 \ | + | 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>xy + yz + zx \ | + | 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 === | ||
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<center> | <center> | ||
<math> | <math> | ||
− | \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \ | + | \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \geq \frac{9}{2} |
</math>, | </math>, | ||
</center> | </center> | ||
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We may normalize so that <math>a+b+c =1 </math>. | We may normalize so that <math>a+b+c =1 </math>. | ||
− | We first note that by the [[rearrangement inequality]] or the fact that <math>(a-b)^2 + (b-c)^2 + (c-a)^2 \ | + | 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) \ | + | 3 (ab + bc + ca) \leq a^2 + b^2 + c^2 + 2(ab + bc + ca) |
</math>, | </math>, | ||
</center> | </center> | ||
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<center> | <center> | ||
<math> | <math> | ||
− | \frac{1}{a(b+c) + b(c+a) + c(a+b)} \ | + | \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> | ||
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<center> | <center> | ||
<math> | <math> | ||
− | a\cdot \frac{1}{b+c} + b \cdot \frac{1}{c+a} + c \cdot \frac{1}{a+b} \ | + | 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> | ||
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===By Muirhead's and Cauchy=== | ===By Muirhead's and Cauchy=== | ||
− | By Cauchy, <cmath>\sum_{\text{cyc}}\frac{a^2}{ab + ac} \ | + | 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: | + | [[Category:Algebra]] |
− | [[Category: | + | [[Category:Inequalities]] |
Latest revision as of 16:21, 24 October 2022
Nesbitt's Inequality is a theorem which, although rarely cited, has many instructive proofs. It states that for positive ,
,
with equality when all the variables are equal.
All of the proofs below generalize to prove the following more general inequality.
If are positive and
, then
,
or equivalently
,
with equality when all the are equal.
Contents
Proofs
By Smoothing
We may normalize so that . Then we wish to show
The equality case is when
so we can use smoothing. We wish to show that if
then
If this is true, then if the minimum occurs at a point where
then we can replace
and
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 which is true because
.
By Rearrangement
Note that and
,
,
are sorted in the same order. Then by the rearrangement inequality,
.
For equality to occur, since we changed to
, we must have
, so by symmetry, all the variables must be equal.
By Cauchy
By the Cauchy-Schwarz Inequality, we have
,
or
,
as desired. Equality occurs when , i.e., when
.
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
,
which yields the desired inequality.
By Expansion and AM-GM
We consider the equivalent inequality
.
Setting , we expand the left side to obtain
,
which follows from , etc., by AM-GM, with equality when
.
By AM-HM
The AM-HM inequality for three variables,
,
is equivalent to
.
Setting yields the desired inequality.
By Substitution
The numbers satisfy the condition
. Thus it is sufficient to prove that if any numbers
satisfy
, then
.
Suppose, on the contrary, that . We then have
, and
. Adding these inequalities yields
, a contradiction.
By Normalization and AM-HM
We may normalize so that . It is then sufficient to prove
,
which follows from AM-HM.
By Weighted AM-HM
We may normalize so that .
We first note that by the rearrangement inequality or the fact that ,
,
so
.
Since , weighted AM-HM gives us
.
By Muirhead's and Cauchy
By Cauchy,
since
by Muirhead as
Another Interesting Method
Let
And
And
Now, we get
Also by AM-GM;
and
By Muirhead's and expansion
Let . Expanding out we get that our inequality is equivalent to
This means
So it follows that we must prove
So it follows that we must prove
which immediately follows from Muirhead's.