How would you show the inequality

\(\displaystyle 1+\frac{1}{n}(n-m) \geq (n-m)m \geq m^2\)

for natural number \(\displaystyle n,m\) satisfying \(\displaystyle n\geq2m\)?

The right inequality is trivial but the left one is false. Take \(\displaystyle m=1\,,\,n=3\):

\(\displaystyle 1+\frac{1}{3}(3-1)\geq (3-1)\cdot 1\iff 1+\frac{2}{3}\geq 2\) , which is false. You may want to try also with, say \(\displaystyle m=2\,,\,n=6\)

It's pretty clear: \(\displaystyle \frac{1}{n}(n-m)<1\) always, whereas we can make the right hand grater than 2 easily.

Tonio