TITLE: Is it valid to use operations on both sides before inequality is proven?
QUESTION [9 upvotes]: 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.

REPLY [2 votes]: Other answers correct the form of this proof by demonstrating that the steps of the proof also imply each other in the reverse order. One could also correct this proof by converting it into a proof by contradiction and maintaining the order of steps.
As noted in the comments, the proposition is not true when $a = b$. Below I show that $ \frac{ab}{a^2+b^2} \leq 1 / 2$.
$$\frac{ab}{a^2+b^2} > 1 / 2 \implies 2ab > a^2 + b^2 \implies 0 > (a-b)^2$$
As you noted, $x^2 \geq 0$ for all real $x$, so the original assumption cannot be true.
I feel that this presentation improves the readability of the proof because it is read in the same order that it was most likely discovered.