\(P+1=x^2+y^2+z^2+xy+yz+2zx=\left(x+z\right)^2+y\left(x+z\right)+y^2=\left(x+z+\dfrac{y}{2}\right)^2+\dfrac{3}{4}y^2\ge0\)
\(\Rightarrow P\ge-1\)
\(P=-1\) khi \(x=-z=\pm\dfrac{1}{\sqrt{2}};y=0\)
Lại có:
\(\left(\sqrt{3}+1\right)\left(x^2+z^2\right)\ge2\left(\sqrt{3}+1\right)xz\)
\(x^2+\dfrac{2+\sqrt{3}}{2}y^2\ge\left(\sqrt{3}+1\right)xy\)
\(z^2+\dfrac{2+\sqrt{3}}{2}y^2\ge\left(\sqrt{3}+1\right)yz\)
Cộng vế:
\(\left(2+\sqrt{3}\right)\left(x^2+y^2+z^2\right)\ge\left(\sqrt{3}+1\right)\left(xy+yz+2zx\right)\)
\(\Rightarrow xy+yz+2zx\le\dfrac{2+\sqrt{3}}{\sqrt{3}+1}=\dfrac{1+\sqrt{3}}{2}\)
Dấu "=" xảy ra khi \(x=z=\pm\sqrt{\dfrac{3+\sqrt{3}}{12}};y=\pm\sqrt{\dfrac{3-\sqrt{3}}{6}}\)
