\(S^2=\left(\left|x\right|+\left|y\right|+\left|x\right|\right)^2=x^2+y^2+z^2+2\left(\left|x\right|\left|y\right|+\left|y\right|\left|z\right|+\left|z\right|\left|x\right|\right)\)
\(S^2=x^2+y^2+z^2+\left|x\right|\left(\left|y\right|+\left|z\right|\right)+\left|y\right|\left(\left|z\right|+\left|x\right|\right)+\left|z\right|\left(\left|x\right|+\left|y\right|\right)\)
Áp dụng BĐT chứa dấu GTTĐ ta có:
\(\left|y\right|+\left|z\right|\ge\left|y+z\right|=\left|-x\right|=\left|x\right|\Rightarrow\left|x\right|\left(\left|y\right|+\left|z\right|\right)\ge z^2\)
Cmtt:\(\left|y\right|\left(\left|z\right|+\left|x\right|\right)\ge y^2,\left|z\right|\left(\left|x\right|+\left|y\right|\right)\ge z^2\)
Vì vậy \(S^2\ge2\left(x^2+y^2+z^2\right)\Rightarrow S^2\ge16\Rightarrow S\ge4\)
Dấu "=" xảy ra khi (x;y;z)=(2;-2;0) và hoán vị của nó, ta có S=4