Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1974 USAMO Problems/Problem 2"
(added solution 6)
Line 67: Line 67:
  
 
J.Z.
 
J.Z.
 +
 +
==Solution 6==
 +
 +
Cubing both sides of the given inequality gives <cmath>a^{3a}b^{3b}c^{3c}\ge(abc)^{a+b+c}</cmath>
 +
If we take <math>a^{3a}b^{3b}c^{3c}</math> as the product of <math>3a</math> <math>a</math>'s, <math>3b</math> <math>b</math>'s, and <math>3c</math> <math>c's</math>, we get that
 +
 +
<center>
 +
<math>\sqrt[3(a+b+c)]{a^{3a}b^{3b}c^{3c}}\ge\frac{3(a+b+c)}{\frac{3a}{a}+\frac{3b}{b}+\frac{3c}{c}}=\frac{a+b+c}{3}\ge\sqrt[3]{abc}</math>
 +
</center>
 +
 +
by GM-HM, as desired.
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Revision as of 19:04, 14 March 2019

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.

Solution 3

WLOG let $a\ge b\ge c$. Let $b = ax$ and $c = ay$, where $x \ge 1$ and $y \ge 1$.

We want to prove that $(a)^{a}(ax)^{ax}(ay)^{ay} \ge (a \cdot ax \cdot ay)^{\frac{a + ax + ay}{3}}$.

Simplifying and combining terms on each side, we get $a^{a + ax + ay}x^{ax}y^{ay} \ge a^{a + ax + ay}(xy)^{\frac{a + ax + ay}{3}}$.

Since $a > 0$, we can divide out $a^{a + ax + ay}$ to get $x^{ax}y^{ay} \ge (xy)^{\frac{a + ax + ay}{3}}$.

Take the $a$th root of each side and then cube both sides to get $x^{3x}y^{3y} \ge (xy)^{1 + x + y}$.

This simplifies to $x^{2x-1}y^{2y-1} \ge x^{y}y^{x}$.

Since $2x - 1 \ge x$ and $2y - 1 \ge y$, we only need to prove $x^{x}y^{y} \ge x^{y}y^{x}$ for our given $x, y$.

WLOG, let $y \ge x$ and $y =kx$ for $k \ge 1$. Then our expression becomes

$x^{x}(xk)^{xk} \ge x^{xk}(xk)^{x}$

$x^{x+xk}k^{xk} \ge x^{x+xk}k^{x}$

$k^{xk} \ge k^{x}$

$k^k \ge k$

This is clearly true for $k \ge 1$.

Solution 4

WLOG let $a\ge b\ge c$. Then sequence $(a,b,c)$ majorizes $(\frac{a+b+c}{3},\frac{a+b+c}{3},\frac{a+b+c}{3})$. Thus by Muirhead's Inequality, we have $\sum_{sym} a^ab^bc^c \ge \sum_{sym} a^{\frac{a+b+c}{3}}b^{\frac{a+b+c}{3}}c^{\frac{a+b+c}{3}}$, so $a^ab^bc^c \ge (abc)^{\frac{a+b+c}{3}}$.

Solution 5

Let $x=\frac{a}{\sqrt[3]{abc}},$ $y=\frac{b}{\sqrt[3]{abc}}$ and $z=\frac{c}{\sqrt[3]{abc}}.$ Then $xyz=1$ and a straightforward calculation reduces the problem to \[x^xy^yz^z \ge 1.\] WLOG, assume $x\ge y\ge z.$ Then $x\ge 1,$ $z\le 1$ and $xy=\frac{1}{z} \ge 1.$ Therefore, \[x^xy^yz^z=x^{x-y}(xy)^{y-z}(xyz)^z \ge 1.\]

J.Z.

Solution 6

Cubing both sides of the given inequality gives \[a^{3a}b^{3b}c^{3c}\ge(abc)^{a+b+c}\] If we take $a^{3a}b^{3b}c^{3c}$ as the product of $3a$ $a$'s, $3b$ $b$'s, and $3c$ $c's$, we get that

$\sqrt[3(a+b+c)]{a^{3a}b^{3b}c^{3c}}\ge\frac{3(a+b+c)}{\frac{3a}{a}+\frac{3b}{b}+\frac{3c}{c}}=\frac{a+b+c}{3}\ge\sqrt[3]{abc}$

by GM-HM, 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1974_USAMO_Problems/Problem_2&oldid=104343"