<s>[INST] 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.
 [/INST] Notice that:
$$(a^2+b^2)^2 = a^4 + b^4 + 2a^2b^2$$
So, then we have that: $(a^2 + b^2)^2 - 2a^2b^2 + 2 \ge 4ab$
It follows that:
$$(a^2 + b^2)^2 \ge 2a^2b^2 + 4ab +2 -4,\ (a^2 + b^2)^2 \ge 2(ab+1)^2 -2^2 = \left(\sqrt{2}(ab+1) + 2\right)\left(\sqrt{2}(ab+1) - 2\right)$$
Our inequality now becomes:
$$(a^2 + b^2)^2 \ge\left(\sqrt{2}ab+\sqrt{2} + 2\right)\left(\sqrt{2}ab+\sqrt{2} - 2\right)$$
NOTE This manipulation may or may not assist you in proving the inequality. I only provided it as a possible way to help you in your proof.
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