Prove Inequality: (a+2)(b+2) ≥ cd with a^2 + b^2 + c^2 + d^2 = 4

[anemone]

Here is this week's POTW:

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Let $a$, $b$, $c$ and $d$ be real numbers with $a^2 + b^2 + c^2 + d^2 = 4$.

Prove that the inequality $(a+2)(b+2) \ge cd$ holds.

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[anemone]

No one answered last week's POTW. You can read the solution from other as follows:

Spoiler

$\begin{align*}0 &\leq (2 + a + b)^2 \\&= 4 + 4(a + b) + (a + b)^2 \\&= 8 + 4a + 4b + 2ab + a^2 + b^2 – 4 \\&= 2(2 + a)(2 + b) – c^2 – d^2 \\&\leq 2(2 + a)(2 + b) – 2cd \end{align*}$

$\implies (a+2)(b+2) \ge cd \,\,\text{(Q.E.D.)}$