Đặt $\dfrac 1x=a; \dfrac 1b=y;\dfrac 1c=z(x;y;z>0).$ Khi đó:
$x+y+z=1$
$\to \left(\dfrac 1a+\dfrac 1b\right)abc=\dfrac{x+y}{xyz}$
$\to A\geqslant \dfrac{4x+4y}{(x+y)^2+z}=\dfrac 4{(x+y)z}\geqslant \dfrac{16}{(x+y+z)^2}=16$
Dấu $"="$ xảy ra khi $a=b=4; c=2$