\(T\ge\frac{x^2}{\sqrt{2\left(y^2+z^2\right)}}+\frac{y^2}{\sqrt{2\left(x^2+z^2\right)}}+\frac{z^2}{\sqrt{2\left(x^2+y^2\right)}}\)
Đặt \(\left(\sqrt{y^2+z^2};\sqrt{x^2+z^2};\sqrt{x^2+y^2}\right)=\left(a;b;c\right)\Rightarrow a+b=c=2014\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=\frac{b^2+c^2-a^2}{2}\\y^2=\frac{a^2+c^2-b^2}{2}\\z^2=\frac{a^2+b^2-c^2}{2}\end{matrix}\right.\)
\(\Rightarrow T.2\sqrt{2}\ge\frac{b^2+c^2-a^2}{a}+\frac{a^2+c^2-b^2}{b}+\frac{a^2+b^2-c^2}{c}\)
\(T.2\sqrt{2}\ge\frac{\left(b+c\right)^2}{2a}+\frac{\left(a+c\right)^2}{2b}+\frac{\left(a+b\right)^2}{2c}-\left(a+b+c\right)\)
\(T.2\sqrt{2}\ge\frac{4\left(a+b+c\right)^2}{2\left(a+b+c\right)}-\left(a+b+c\right)=a+b+c=2014\)
\(\Rightarrow T\ge\frac{1007}{\sqrt{2}}\)
Dấu "=" xảy ra khi \(x=y=z=...\)