Proving a quadratic inequality Given that $(x+y)^2 \geq 4xy$, then prove $1/x^2 + 1/y^2 \geq 4/(x^2 + y^2)$.
I have tried taking the reciprocal, squaring, square rooting, expanding, factoring and various other algebraic manipulation but nothing has worked so far. 
The question seems to be structured in a way which suggests that I should be manipulating the first inequality (which is already given) to produce the second inequality. Does anyone know how to do this

We can not  make it because it's wrong! Try $x=0$.
The condition gives $$y^2\geq0,$$ which is true, but the statement, which we need to prove is wrong 
because we can not  divide by $0$.
For $xy>0$ we obtain:
$$(x+y)^2\geq4xy$$ it's $$\frac{x^2+2xy+y^2}{xy}\geq4$$ or
$$\frac{x}{y}+1+\frac{y}{x}+1\geq4$$ or
$$(x+y)\left(\frac{1}{x}+\frac{1}{y}\right)\geq4.$$
Now, after replacing $x$ on $x^2$ and $y$ on $y^2$ we obtain:
$$(x^2+y^2)\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\geq4$$ or 
$$\frac{1}{x^2}+\frac{1}{y^2}\geq\frac{4}{x^2+y^2}.$$
