Let a,b $\in$ $\mathbb{R}.$ Show that $a^4+b^4+8\ge 8ab.$ 
Let $a,b \in \mathbb{R}.$ Show that $a^4+b^4+8\ge 8ab.$

The question is from the inequalities section of An Excursion in mathematics by Bhaskaraycharya Pratisthanan. My heuristics include using the AM-GM inequality. I am unable to design the problem to proceed further.
Please provide me some hints.

You're on right path just a small trick using AM-GM
Write $$a^4+b^4+8=a^4+b^4+4+4$$
And then apply AM-GM
$$\frac{a^4+b^4+4+4}{4}\ge \sqrt[4]{16a^4b^4}$$
$$\frac{a^4+b^4+4+4}{4}\ge 2ab$$
$${a^4+b^4+8}\ge 8ab$$
Or You can just SOS it
$$(a^2-b^2)^2\ge0\implies a^4+b^4-2a^2b^2\ge0$$
$$2(ab-2)^2\ge0\implies2a^2b^2+8-8ab\ge0$$
Add above two inequalities to get
$$ a^4+b^4+8\ge8ab$$
