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?
 A: 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 $b$, $\,\, ab \le (1/2)(a^2+b^2)$
A: 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!
A: Since the inequality is homogenious, you can suppose $x+y=1$. The resulting inequality is the $AM-GM$. 

Edit: An inequality $f(x_1,x_2,...,x_n)\ge 0$ is homogenius if there is some $n$ such that for all $f(tx_1,tx_2,...,tx_n)=t^nf(x_1,x_2,...,x_n)$ for all $t>0$. If an inequality is homogenius, you can suppose some condition of normality. For example $x_1+x_2+...+x_n=1$ or $x_1\times x_2\times...\times x_n=1$.

AM-GM is: for $x_1,...,x_n>0$ we have $\frac{x_1+...+x_n}{n}\ge\sqrt[n]{x_1\times...\times x_n}$
A: 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. 
