1974 USAMO Problems/Problem 2

Revision as of 19:04, 14 March 2019 by Thecooldinosuar (talk | contribs) (added solution 6)


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

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



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


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, \[x^xy^yz^z=x^{x-y}(xy)^{y-z}(xyz)^z \ge 1.\]


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


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

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