# Show that $\frac{j(j-1)}{2n}> \frac{j^2}{4nr}$

In Lemmas 8.5 and 8.6 in book Irrational Numbers by Ivan M. Niven it uses the following :

$$\frac{j(j-1)}{2n}> \dfrac{j^2}{4nr}$$

$$n \ge 2$$, $$2 \le j \le n$$ and $$r \ge 2$$, that's it! How the mentioned inequality holds?

The inequality is equivalent to $$\frac{j-1}{j} > \frac{1}{2r}$$ and that is true because for $$j\ge 2$$ $$\frac{j-1}{j} = 1-\frac 1j \ge \frac 12$$ and for $$r \ge 2$$ $$\frac{1}{2r} \le \frac 14 \, .$$
For the given range, $$j-1\ge\frac j2$$.
Multiply by $$j/2n\implies\frac{j(j-1)}{2n}\ge\frac{j^2}{4n}>\frac{j^2}{4nr}$$.