Giải thích các bước giải:
Ta có:
$A=\dfrac{1}{xy}+\dfrac{1}{xz}$
$\to A=\dfrac1x(\dfrac1y+\dfrac1z)$
$\to A\ge \dfrac1x\cdot\dfrac{4}{y+z}$
$\to A\ge \dfrac{4}{x(y+z)}$
Mà $x(y+z)\le \dfrac14(x+y+z)^2=4$ vì $x+y+z=4$
$\to \dfrac{4}{x(y+z)}\ge \dfrac44=1$
$\to \dfrac{1}{xy}+\dfrac{1}{xz}\ge 1$