<s>[INST] 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?
 [/INST] Since $x$ and $y$ are positive, so is $x+y$, and $0\leq(x-y)^2$ since squares are always non-negative. So, $0\leq (x+y)(x-y)^2$, which expands to $0\leq x^3+y^3-xy^2-x^2y$. We can rewrite this to $$(x+y)\cdot xy \leq \tfrac{1}{2}(x(x^2+y^2)+y(x^2+y^2))$$ and dividing by $xy\cdot (x^2+y^2)$ yields
$$\frac{(x+y)\cdot xy}{xy(x^2+y^2)}=\frac{x+y}{x^2+y^2} \leq \tfrac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)=\tfrac{1}{2}\frac{x+y}{xy}=\tfrac{1}{2}\frac{x(x^2+y^2)+y(x^2+y^2)}{xy(x^2+y^2)}$$
$$\frac{x+y}{x^2+y^2}\leq\tfrac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)$$ which is the wanted inequality.
Hope this helped!
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