IMO 1996 SL A1

Origin: SLO | Category: Algebra

Problem

Let $a$, $b$, and $c$ be positive real numbers such that $abc=1$.

Prove that

$$\frac{ab}{a^5+b^5+ab} + \frac{bc}{b^5+c^5+bc} + \frac{ca}{c^5+a^5+ca} \leq 1.$$

When does equality hold?

Solution

We have

$$a^5+b^5-a^2b^2(a+b) = (a^3-b^3)(a^2-b^2) \geq 0,$$

i.e.

$$a^5+b^5 \geq a^2b^2(a+b).$$

Hence

$$\frac{ab}{a^5+b^5+ab} \leq \frac{ab}{a^2b^2(a+b)+ab} = \frac{abc^2}{a^2b^2c^2(a+b)+abc^2} = \frac{c}{a+b+c}.$$

Now, the left-hand side of the inequality to be proved does not exceed

$$\frac{c}{a+b+c} + \frac{a}{a+b+c} + \frac{b}{a+b+c} = 1.$$

Equality holds if and only if $a=b=c$.