\(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8xyz\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)-8xyz\ge0\)
Ta có: \(x+y\ge2\sqrt{xy}\)
\(y+z\ge2\sqrt{yz}\)
\(x+z\ge2\sqrt{xz}\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8\sqrt{x^2y^2z^2}\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8\left|x\right|\left|y\right|\left|z\right|\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge8xyz\)
Áp dụng bất đẳng thức cô - si , ta có :
\(\frac{x+y}{2}\ge\sqrt{xy}\Rightarrow x+y\ge2\sqrt{xy}\left(1\right)\)
\(\frac{y+z}{2}\ge\sqrt{yz}\Rightarrow y+z\ge2\sqrt{yz}\left(2\right)\)
\(\frac{z+x}{2}\ge\sqrt{zx}\Rightarrow z+x\ge2\sqrt{zx}\left(3\right)\)
Từ \(\left(1\right);\left(2\right);\left(3\right)\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge2\sqrt{xy}2\sqrt{yz}2\sqrt{zx}\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8\left(\sqrt{xyz}\right)^2\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8xyz\left(đpcm\right)\)
