<s>[INST] 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 ?
 [/INST] $$a^4+b^4+a^2b^2=(a^2+b^2)^2-a^2b^2=60$$
$$(a^2+b^2-ab)(a^2+b^2+ab)=60$$
Solving equations $$a^2+b^2-ab=6$$$$a^2+b^2+ab=10$$ will give you the result.
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