Ta có: \(2\sqrt{2}-1< \sqrt{5}\)
\(A^2=x+y+z+2\left(\sqrt{xy}+yz+zx\right)=5+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\ge5\)
\(\Rightarrow A\ge\sqrt{5}>2\sqrt{2}-1\Rightarrow A-2\sqrt{2}+1>0\)
\(0\le x;y;z\le2\Rightarrow0\le\sqrt{x};\sqrt{y};\sqrt{z}\le\sqrt{2}\)
\(\Rightarrow\left(\sqrt{x}-\sqrt{2}\right)\left(\sqrt{y}-\sqrt{2}\right)+\left(\sqrt{y}-\sqrt{2}\right)\left(\sqrt{z}-\sqrt{2}\right)+\left(\sqrt{x}-\sqrt{2}\right)\left(\sqrt{z}-\sqrt{2}\right)\ge0\)
\(\Leftrightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\ge2\sqrt{2}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)-6=2\sqrt{2}A-6\)
\(\Rightarrow A^2\ge5+2\left(2\sqrt{2}A-6\right)\)
\(\Leftrightarrow A^2-4\sqrt{2}A+7\ge0\)
\(\Leftrightarrow\left(A-2\sqrt{2}+1\right)\left(A-2\sqrt{2}-1\right)\ge0\)
\(\Leftrightarrow A-2\sqrt{2}-1\ge0\)
\(\Rightarrow A\ge2\sqrt{2}+1\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;2;2\right)\) và hoán vị
