How to prove this: $a^4+b^4+2 \ge 4ab$? How to prove this: $a^4+b^4+2 \ge 4ab$?
$a$ and $b$ are reals.

Inequalities like this can most often be proved or simplified by using the Arithmetic-Geometric mean inequality.
First apply AM-GM in $a^4$ and $b^4$:
$$\frac{a^4+b^4}{2} \ge \sqrt{a^4b^4}$$
$$a^4+b^4 \ge 2a^2b^2$$
Now apply AM-GM in $2a^2b^2$ and $2$:
$$\frac{2a^2b^2+2}{2} \ge \sqrt{4a^2b^2}$$ 
$$2a^2b^2+2 \ge 4ab$$
I suppose the result is quite obvious now. This answer was, of course, posted with the assumption that $a$ and $b$ are reals.
