If $a^2+b^2+c^2+d^2=4$ then $(a+2)(b+2)\geq cd$

Question

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?

Answer 1

Because by AM-GM $$(a+2)(b+2)=ab+2(a+b)+4=$$ $$=\frac{1}{2}(2ab+4a+4b+4+a^2+b^2)+2-\frac{1}{2}(a^2+b^2)=$$ $$=\frac{1}{2}(a+b+2)^2+2-\frac{1}{2}(a^2+b^2)\geq$$ $$\geq2-\frac{1}{2}(a^2+b^2)=\frac{1}{2}(c^2+d^2)\geq |cd|\geq cd.$$

Answer 2

Using the AM-GM inequality, we have $$(a+2)(b+2) = \frac{4-a^2-b^2-c^2-d^2}{2}+\frac{(a+b+2)^2}{2}+\frac{c^2+d^2}{2}$$ $$\geqslant \frac{c^2+d^2}{2} \geqslant cd.$$