<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] Just notice that the following inequalities are equivalent to each other:
$$
\begin{align*}
\frac {x+y}{x^2+y^2} &\leq  \frac 12 \left(\frac {1}{x}+\frac{1}{y}\right)\\
\frac {x+y}{x^2+y^2} &\le \frac12 \cdot \frac{x+y}{xy}\\
\frac 1{x^2+y^2} &\le \frac12 \cdot \frac{1}{xy}\\
2xy &\le x^2+y^2
\end{align*}
$$
The last one is a well-known inequality:


*

*Simple algebra question - proving $a^2+b^2 \geqslant 2ab$

*Show that $2 xy < x^2 + y^2$ for $x$ is not equal to $y$

*Prove the inequality $|xy|\leq\frac{1}{2}(x^2+y^2)$

*Show that for all real numbers $a$ and $b$, $\,\, ab \le (1/2)(a^2+b^2)$
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