Với $0<x\leq \frac{\pi}{3}$. Đặt $t=tanx$ thì $0<t\leq \sqrt{3}$Ta có: $\frac{cosx}{sin^2x(2cosx-sinx)}=\frac{\frac{cosx}{cos^3x}}{\frac{sin^2x}{cos^3x}(2cosx-sinx)}$
$=\frac{t^2+1}{t^2(2-t)}=\frac{(t-1)^2+2t}{t^2(2-t)}\geq \frac{2t}{t^2(2-t)}=\frac{2}{1-(t-1)^2}\geq 2$