if ${m/n}$ is an approximation to $\sqrt{2}$, prove $m/2n+n/m$ is always a better approximation.

Question

I noticed $(m/2n+n/m)^2 -2$ = $(m/2n-n/m)^2$.

So I set $(m/n)^2 - 2 = d$

and substituted value for n in the above equation but can't show that resulting equation is less than d.