1974 USAMO Problems/Problem 2

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.

Solution 7

Replacing $a$ with $ak$ , $b$ with $bk$ , and $c$ with $ck$ , for some positive real $k$ we get: $a^{ak}k^{ak}b^{bk}k^{bk}c^{ck}k^{ck} = a^{ak}b^{bk}c^{ck}k^{ak+bk+ck} \ge {k^3abc}^{{ak+bk+ck}/3} = k^{ak+bk+ck}{abc}^{{ak+bk+ck}/3}$ This means that this inequality is homogeneous since both sides have the same power of $k$ as a factor. Since the inequality is homogeneous, we can scale $abc$ so that their product is $1$ , i.e. $abc = 1$ . This makes the inequality turn into something much more nicer to deal with. Now we have to prove: $a^ab^bc^c \ge 1$ given that $abc = 1$ . Note that $a^a \ge a$ even if $a \le 1$ . Therefore $a^a \ge a$ , $b^b \ge b$ , and $c^c \ge c$ . Multiplying these together we get: $a^ab^bc^c \ge abc = 1$ . This proves the desired result. Equality holds when $a = b = c = 1$ .

Solution 8 (Rearrangement)

Let $\log_2a=x$ , $\log_2b=y$ and $\log_2c=z$ , and WLOG $a\ge b\ge c >0$ . Then we have both $a\ge b\ge c$ and $x\ge y \ge z$ . By the rearrangement inequality, $ax+by+cz\ge ay + bz +cx$ and $ax+by +cz\ge az+bx+cy$ . Summing, $2(ax+by+cz)\ge a(y+z)+b(x+z)+c(x+y)$ Adding $ax+by+cz$ ,we get $ax+by+cz\ge \frac{(a+b+c)(x+y+z)}{3}$ . Now we substitute back for $x,y,z$ to get: $\log_2a^ab^bc^c\ge \log_2(abc)^{\frac{a+b+c}{3}}$ . Raising $2$ to the power of each side, we get $a^ab^bc^c\ge (abc)^{\frac{a+b+c}{3}}$

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

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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