It's a fact that:
$\frac{a+b}{2} \ge \sqrt{ab}$
When a and b are non negative real numbers. What's the error in the following demonstration:
$\frac{a+b}{2} \ge \sqrt{ab}$ , then
$a+b \ge 2\sqrt{ab}$ , then
$a^{2}+2ab+b^{2} \ge 4ab$ , then
$a^{2}-2ab+b^{2} \ge 0$ , then
$(a-b)^{2} \ge 0$
discrete math professor gave that question as homework, and this is a direct translation from portuguese.
the only idea I have is to say that both $(a-b)^{2}\ge 0$ and $(b-a)²\ge 0$ are valid, but it only specifies one. Is that correct?