Theo BĐT Bunhiacopski:
$1=\sqrt{x}.\sqrt{y}+
\sqrt{y}.\sqrt{z} +
\sqrt{z}.\sqrt{x} \leq \sqrt{(x+y+z)(y+z+x)} =x+y+z$
$\left ( x +y+z\right ) ^{2}=\left ( \frac{x}{\sqrt{x+y}}
\sqrt{x+y} +
\frac{y}{\sqrt{y+z}} \sqrt{y+z}+
\frac{z}{\sqrt{z+x}} \sqrt{z+x} \right)^{2} $
$\leq T.\left ( x+y+y+z+z+x \right )=2T
\left ( x +y+z\right )$
$ \Rightarrow T\geq \frac{1}{2}
\left ( x +y+z\right ) \geq
\frac{1}{2} $
Dấu "=" xảy ra $\Leftrightarrow x=y=z=\frac{1}{3}$
Vậy $Min(T)=
\frac{1}{2} $ khi $
x=y=z=\frac{1}{3}$