$T=\dfrac{1}{x}+\dfrac{25}{y}+\dfrac{64}{z}$
$⇒49.T=(4x+9y+16z).\left (\dfrac{1}{x}+\dfrac{25}{y}+\dfrac{64}{z} \right )$
$=[(2\sqrt{x})^2+(3\sqrt{y})^2+(4\sqrt{z})^2].\left [\left (\dfrac{1}{\sqrt{x}} \right )^2+\left (\dfrac{5}{\sqrt{y}} \right )^2+\left (\dfrac{8}{\sqrt{z}} \right ) \right ]$
Áp dụng BĐT Bunhiacopxki vào $6$ số $2\sqrt{x} ; 3\sqrt{y} ; 4\sqrt{z}$ và $\dfrac{1}{\sqrt{x}} ; \dfrac{5}{\sqrt{y}} ; \dfrac{8}{\sqrt{z}} :$
$49T=[(2\sqrt{x})^2+(3\sqrt{y})^2+(4\sqrt{z})^2].\left [\left (\dfrac{1}{\sqrt{x}} \right )^2+\left (\dfrac{5}{\sqrt{y}} \right )^2+\left (\dfrac{8}{\sqrt{z}} \right ) \right ] \geq \left (2\sqrt{x}.\dfrac{1}{\sqrt{x}}+3\sqrt{y}.\dfrac{5}{\sqrt{y}}+4\sqrt{x}.\dfrac{8}{\sqrt{z}} \right )^2=(2+15+32)^2=49^2$
$⇒49T \geq 49^2$
Hay $T \geq 49$
Dấu "=" xảy ra khi :
$\left\{ \begin{matrix}\dfrac{1}{2x}=\dfrac{5}{3y}=\dfrac{8}{4z}\\4x+9y+16z=49\end{matrix} \right.$
$⇔\left\{ \begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{5}{3}\\z=2\end{matrix} \right.$
