Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that : Let $a^{4} + b^{4} + a^{2} b^{2} = 60 \ \ a,b \in \mathbb{R}$ Then prove that :
$$4a^{2} + 4b^{2} - ab \geq 30$$

My attempt: :
$$4a^{2} + 4b^{2} - ab \geq 30 \\ 4(a^2+b^2)-ab \geq30 \\4(60-a^2b^2)-ab\geq30\\ 240-30\geq4(ab)^2+ab\\ 4(ab)^2+ab \leq210$$
Now what ?

Let $a^2+b^2=2kab$.
Hence, we need to prove that
$$(4a^2+4b^2-ab)^2\geq900\cdot\frac{a^4+b^4+a^2b^2}{60}$$ or
$$(8k-1)^2\geq15(4k^2-1)$$ or
$$(k-2)^2\geq0.$$
Done!
