WakeUp wrote: Let $x$ and $y$ be two positive reals. Prove that $xy\le\frac{x^{n+2}+y^{n+2}}{x^n+y^n}$ for all non-negative integers $n$. you can use Chebyshev inequality We have $(x^{n+2}+y^{n+2}\geq \frac{1}{2}(x^{n}+y^{n})(x^{2}+y^{2})\geq xy(x^{n}+y^{n})$ Therefore $ xy\le\frac{x^{n+2}+y^{n+2}}{x^{n}+y^{n}} $ Complete prove

$ 0\leq(x-y)(x^{n+1}-y^{n+1})=(x^{n+2}+y^{n+2})-xy(x^{n}+y^{n}) $

Notice that $ xy(x^{n} + y^{n}) = x^{n+1}y + xy^{n+1} $ Let's assume wlog that $ x\ge y $. By rearrangement inequality we have that $ x^{n+2} + y^{n+2}\ge x^{n+1}y + xy^{n+1} $. That completes it.

WakeUp wrote: Too easy to be in a mathematical competition surely... Quote: Pan African Olympiad 2008