Is it valid to use operations on both sides before inequality is proven? As part of a bigger proof, I am trying to prove the inequality:
$$\frac{ab}{a^2 + b^2}< \frac{1}{2}$$
Is the following proving method correct?
$$\frac{ab}{(a^2)+(b^2)} <\frac{1}{2}$$
$$2ab < (a^2 + b^2)$$
$$0 <(a-b)^2$$
$a \neq b$ thus $a - b \neq 0$ and every number squared is not negative thus $(a-b)^2$ is positive thus bigger than zero
My concern with this proof is that I am not sure if I am allowed to do operations on both sides before inequality is proven. 

What you've shown is $$\frac{ab}{a^2 + b^2} < \frac{1}{2} \implies 0 < (a-b)^2$$
But what you are (apparently) required to show is $$0 < (a-b)^2 \implies \frac{ab}{a^2 + b^2} < \frac{1}{2}$$
So the correct proof is basically the reverse of what you've written, namely:
Suppose $a ≠ b$. Then 
$$\begin{align} 0 < (a-b)^2  &\implies  2ab < (a^2 + b^2)\\
& \implies\frac{ab}{(a^2)+(b^2)} <\frac{1}{2}
\end{align}$$
$\square$
