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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?

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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 $r \ge 2$ $$ \frac{1}{2r} \le \frac 14 \, . $$

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For the given range, $j-1\ge\frac j2$.

Multiply by $j/2n\implies\frac{j(j-1)}{2n}\ge\frac{j^2}{4n}>\frac{j^2}{4nr}$.

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