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Difference between revisions of "1974 USAMO Problems/Problem 2"
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<center><math>a^ab^bc^c\ge (abc)^{(a+b+c)/3}</math></center>
 
<center><math>a^ab^bc^c\ge (abc)^{(a+b+c)/3}</math></center>
  
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==Solution==
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==Solution 1==
 
Consider the function <math>f(x)=x\ln{x}</math>. <math>f''(x)=\frac{1}{x}>0</math> for <math>x>0</math>; therefore, it is a convex function and we can apply [[Jensen's Inequality]]:
 
Consider the function <math>f(x)=x\ln{x}</math>. <math>f''(x)=\frac{1}{x}>0</math> for <math>x>0</math>; therefore, it is a convex function and we can apply [[Jensen's Inequality]]:
 
<center><math>\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\frac{a+b+c}{3}\right)</math></center>
 
<center><math>\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\frac{a+b+c}{3}\right)</math></center>
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which simplifies to the desired inequality.
 
which simplifies to the desired inequality.
  
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==Solution 2==
 +
Note that <math>(a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3</math>.
  
 +
So if we can prove that <math>a^ab^bc^c\ge a^bb^cc^a</math> and <math>a^ab^bc^c\ge a^cb^ac^b</math>, then we are done.
 +
 +
WLOG let <math>a\ge b\ge c</math>.
 +
 +
Note that <math>(a^ab^bc^c)\cdot \left(\dfrac{c}{a}\right)^{a-b}\cdot \left(\dfrac{c}{b}\right)^{b-c}=a^bb^cc^a</math>. Since <math>\dfrac{c}{a} \le  1</math>, <math>\dfrac{c}{b} \le  1</math>, <math>a-b \ge 0</math>, and <math>b-c \ge 0</math>, it follows that <math>a^ab^bc^c \ge a^bb^cc^a</math>.
 +
 +
Note that <math>(a^ab^bc^c)\cdot \left(\dfrac{b}{a}\right)^{a-b}\cdot \left(\dfrac{c}{a}\right)^{b-c}=a^cb^ac^b</math>. Since <math>\dfrac{b}{a} \le  1</math>, <math>\dfrac{c}{a} \le  1</math>, <math>a-b \ge 0</math>, and <math>b-c \ge 0</math>, it follows that <math>a^ab^bc^c \ge a^cb^ac^b</math>.
 +
 +
Thus, <math>(a^ab^bc^c)^3\ge (a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3</math>, and cube-rooting both sides gives <math>a^ab^bc^c\ge (abc)^{(a+b+c)/3}</math> as desired.
  
  

Revision as of 23:53, 24 April 2014

Problem

Prove that if $a$, $b$, and $c$ are positive real numbers, then

$a^ab^bc^c\ge (abc)^{(a+b+c)/3}$

Solution 1

Consider the function $f(x)=x\ln{x}$. $f''(x)=\frac{1}{x}>0$ for $x>0$; therefore, it is a convex function and we can apply Jensen's Inequality:

$\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\frac{a+b+c}{3}\right)$

Apply AM-GM to get

$\frac{a+b+c}{3}\ge \sqrt[3]{abc}$

which implies

$\frac{a\ln{a}+b\ln{b}+c\ln{c}}{3}\ge \left(\frac{a+b+c}{3}\right)\ln\left(\sqrt[3]{abc}\right)$

Rearranging,

$a\ln{a}+b\ln{b}+c\ln{c}\ge\left(\frac{a+b+c}{3}\right)\ln\left(abc\right)$

Because $f(x) = e^x$ is an increasing function, we can conclude that:

$e^{a\ln{a}+b\ln{b}+c\ln{c}}\ge{e}^{\ln\left(abc\right)(a+b+c)/3}$

which simplifies to the desired inequality.

Solution 2

Note that $(a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3$.

So if we can prove that $a^ab^bc^c\ge a^bb^cc^a$ and $a^ab^bc^c\ge a^cb^ac^b$, then we are done.

WLOG let $a\ge b\ge c$.

Note that $(a^ab^bc^c)\cdot \left(\dfrac{c}{a}\right)^{a-b}\cdot \left(\dfrac{c}{b}\right)^{b-c}=a^bb^cc^a$. Since $\dfrac{c}{a} \le 1$, $\dfrac{c}{b} \le 1$, $a-b \ge 0$, and $b-c \ge 0$, it follows that $a^ab^bc^c \ge a^bb^cc^a$.

Note that $(a^ab^bc^c)\cdot \left(\dfrac{b}{a}\right)^{a-b}\cdot \left(\dfrac{c}{a}\right)^{b-c}=a^cb^ac^b$. Since $\dfrac{b}{a} \le 1$, $\dfrac{c}{a} \le 1$, $a-b \ge 0$, and $b-c \ge 0$, it follows that $a^ab^bc^c \ge a^cb^ac^b$.

Thus, $(a^ab^bc^c)^3\ge (a^ab^bc^c)(a^bb^cc^a)(a^cb^ac^b)=\left((abc)^{(a+b+c)/3}\right)^3$, and cube-rooting both sides gives $a^ab^bc^c\ge (abc)^{(a+b+c)/3}$ as desired.


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

1974 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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