Show that for all real numbers $a$ and $b$, $\,\, ab \le (1/2)(a^2+b^2)$

Remember that for any $x \in \mathbb{R}$, we have that $x^2 \geq 0$ i.e. square of any real number is non-negative.

Hence, $$(a-b)^2 \geq 0$$ since $a,b \in \mathbb{R}$. Expand $(a-b)^2$ and rearrange to get what you want. \begin{align} (a-b)^2 & \geq 0\\ a^2 + b^2 - 2ab & \geq 0\\ a^2 + b^2 & \geq 2ab \end{align}

Hint: $\forall a,b \in \mathbb R, \; (a-b)^2\ge 0$

Remember that any real number squared is non-negative, and

$$a^2+b^2\geq 2ab\Longleftrightarrow a^2-2ab+b^2=(a-b)^2\geq 0\,\,...$$