Proving the inequality $ \frac {x+y}{x^2+y^2}\leq \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right)$ Let $x$ and $y$ be two positive numbers: 
Prove that $$ \left( \frac {x+y}{x^2+y^2}\right) \leq  \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right).$$
I answered this one by squaring the two expressions. And therefore finding the difference after squaring the formulas. I don't even know if it's right but I wanted to find another way to answer this question?

An alternative approach: 
$\left( \frac {x+y}{x^2+y^2}\right) \leq  \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right) \Longleftrightarrow 4x^2y + 4xy^2 \leq (x^2 + y^2)(2y + 2x) \Longleftrightarrow x^2y + y^2x \leq  x^3 + y^3$
We assume WLOG $x \leq y \Longrightarrow \exists c\in \mathbb{R}_{≥0}$ such that $x+c = y$
Hence the rightmost inequality can be simplified to $0 \leq (x^3 + (x+c)^3) -(x^2(x+c) + x(x+c)^2) = c^2(c+2x)$ which is clear. 
