Giải cho $x,y,z>0$,
Áp dụng BĐT Bunhia, ta có:
$(x+y+z)\left(\dfrac{1}{x}+\dfrac{4}{y}+\dfrac{9}{z}\right)\ge(1+2+3)^2=36$
$\Rightarrow \dfrac{1}{x}+\dfrac{4}{y}+\dfrac{9}{z}\ge\dfrac{36}{x+y+z}\ge3$
Dấu bằng xảy ra khi: $\left\{\begin{array}{l}\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}\\x+y+z=12\end{array}\right. \Leftrightarrow \left\{\begin{array}{l}x=2\\y=4\\z=6\end{array}\right.$