Bernoulli’s Inequality

Bernoulli’s Inequality states that for real numbers $x \geq -1$, $r \geq 0$ it holds that

$$ \left(1+x \right)^r \geq 1+xr $$

Problem (USAMO, 1991)

Let

$$ a =\frac{m^{m+1} + n^{n+1}}{m^m + n^n}$$

where $m$ and $n$ are positive integers. Prove that

$$ a^m + a^n \geq m^m + n^n $$

Solution

Let us apply the Bernoulli’s Inequality to $\left( \frac{a}{m} \right)^m$:

$$ \left( \frac{a}{m} \right)^m = \left( 1+\frac{a-m}{m} \right)^m \geq 1+\frac{a-m}{m} \cdot m = 1+a-m $$

which implies that

$$ a^m \geq m^m \left( 1+a-m \right) $$

Let us now apply the Bernoulli’s Inequality to $\left( \frac{a}{n} \right)^n$:

$$\left( \frac{a}{n} \right)^n = \left( 1+\frac{a-n}{n} \right)^n \geq 1+\frac{a-n}{n} \cdot m = 1+a-n $$

which implies that

$$ a^n \geq n^n \left( 1+a-n \right) $$

Therefore

$$ a^m + a^n \geq m^m \left( 1+a-m \right) + n^n \left( 1+a-n \right) = m^m + n^n $$