Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{4}{y}=b\\\dfrac{9}{z}=c\end{matrix}\right.\)
Ta có a+b+c =1 và \(\left\{{}\begin{matrix}\dfrac{1}{a}=x\\\dfrac{4}{b}=y\\\dfrac{9}{c}=z\end{matrix}\right.\)
A= 4x +y +z = \(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{9}{c}\)\(\ge\dfrac{\left(2+2+3\right)^2}{a+b+c}=\dfrac{49}{1}=49\)