từ điều kiện $z \geq x+y \Rightarrow z^2\geq \left ( x+y \right )^2\geq 4xy\Rightarrow z^4\geq 16x^2y^2$$T=3+\frac{x^4}{y^4}+\frac{x^4}{z^4}+\frac{y^4}{x^4}+\frac{y^4}{z^4}+\frac{z^4}{x^4}+\frac{z^4}{y^4}$
$=3+\frac{x^4}{y^4}+\frac{y^4}{x^4}+\frac{x^4}{z^4}+\frac{z^4}{256x^4}+\frac{y^4}{z^4}+\frac{z^4}{256y^4}+\frac{255z^4}{256x^4}+\frac{255z^4}{256y^4}$
Áp Dụng BĐT AM-GM
$T\geq 3+2+2.\sqrt{\frac{1}{256}}+2.\sqrt{\frac{1}{256}}+2.\sqrt{\frac{255^2.z^8}{256^2.x^4.y^4}}$
$T\geq 5+\frac{1}{4}+\frac{2.255.z^4}{256.x^2y^2}\geq 5+\frac{1}{4}+\frac{2.255.16}{256}=\frac{297}{8}$
dấu $''='' \Leftrightarrow x=y=\frac{z}{2}$