Prove inequality $ a^4+b^4 \leq 2 a^2 b^2 +1. $ Let $a,b$  be such that  $|a|\leq 1$, $|b|\leq 1$. Prove that then 
$$
a^4+b^4 \leq 2 a^2 b^2 +1.
$$

Staring at this, I have an uncontrollable urge to complete the square. 
$$a^4 + b^4 \le^{\text{?}} 2a^2 b^2 + 1
$$
$$a^4 - 2a^2 b^2 + b^4 \le^{\text{?}} 1
$$
$$(a^2 - b^2)^2 \le^{\text{?}} 1
$$
Now it's easy. If $|a|, |b| \le 1$, it follows that $0 \le a^2, b^2 \le 1$, and so $|a^2 - b^2| \le 1$ and so $(a^2 - b^2)^2 \le 1$.
