# inequality

#### willy0625

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 $$\displaystyle n\geq2m$$?

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#### tonio

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

willy0625