<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] We have $$[4(a^2+b^2)]^2=16(a^4+b^4+2a^2b^2) =16(60+a^2b^2)$$ 


Then,  $$  [4(a^2+b^2)]^2 - (30 +ab)^2 =16(60+a^2b^2) - (30 +ab)^2 \\=15a^2b^2 -60 ab - 60 =\color{red}{15(ab-2)^2\ge0} $$ 
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