<s>[INST] If $a^2+b^2+c^2+d^2=4$ then $(a+2)(b+2)\geq cd$ Let $a,b,c,d$ be real numbers with $a^2+b^2+c^2+d^2=4$. Prove that $(a+2)(b+2)\geq cd$.
My approach: I have considered an expression $$\begin{aligned}(a+2)(b+2)-cd=&4+2(a+b)+(ab-cd)\\=&(a^2+b^2+c^2+d^2)+2a+2b+(ab-cd)\end{aligned}$$ I was trying to write it as the sum of squares but I failed.
Can anyone show how to solve this problem please?
 [/INST] Using the AM-GM inequality, we have
$$(a+2)(b+2) = \frac{4-a^2-b^2-c^2-d^2}{2}+\frac{(a+b+2)^2}{2}+\frac{c^2+d^2}{2}$$
$$\geqslant \frac{c^2+d^2}{2} \geqslant cd.$$
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