Prove by deduction $\displaystyle \frac{a^2 + b^2}{2} \geq ab$ I am currently working in the following question:

Prove by deduction $$\frac{a^2 + b^2}{2} \geq ab$$ for all $a,b \in \mathbb{R}$

My question is: can prove instead the following?
$$\frac{a^2 + b^2}{2} - ab \geq 0$$
My solution:
$$\frac{a^2 + b^2}{2} - ab = \frac{a^2 + b^2 -2ab}{2} = \frac{(a-b)^2}{2} \geq 0$$

You are correct since $$\frac{a^2+b^2}{2}-ab\geq 0\iff\frac{a^2+b^2}{2}\geq ab.$$
Alternatively since $(a-b)^2\geq0$ for all $a,b\in\mathbb R$ we have $$a^2+b^2\geq2ab$$ or $$\frac{a^2+b^2}{2}\geq ab$$
as required.
