\(P=\sqrt{4-a^2}+\sqrt{4-b^2}+\sqrt{4-c^2}\)
\(\Rightarrow2\sqrt{3}P=\Sigma2\sqrt{3}\sqrt{4-a^2}\)\(=\Sigma2\sqrt{\left(a+b+c\right)\left(4-a^2\right)}\)
Vì \(a,b,c\in\left[-2,2\right]\Rightarrow\) \(\left\{{}\begin{matrix}4-a^2\ge0\\4-b^2\ge0\\4-c^2\ge0\end{matrix}\right.\)
Áp dụng BĐT AM-GM cho các số không âm, ta có:
\(\left(a+b+c\right)+\left(4-a^2\right)\ge2\sqrt{\left(a+b+c\right)\left(4-a^2\right)}\)
\(\Rightarrow2\sqrt{3}P\le\Sigma\left(a+b+c\right)+\left(4-a^2\right)\)
\(\Leftrightarrow2\sqrt{3}P\le3\left(a+b+c\right)+12-\left(a^2+b^2+c^2\right)\)
\(\Rightarrow2\sqrt{3}P\le21-\frac{\left(a+b+c\right)^2}{3}=21-\frac{9}{3}=18\)
\(\Rightarrow P\le3\sqrt{3}\)
\(''=''\Leftrightarrow a=b=c=1\)