Proving inequality for induction proof: $\frac1{(n+1)^2} + \frac1{n+1} < \frac1n$

To prove the inequality you were required to prove, note

$\dfrac1{n+1}+\dfrac1{(n+1)^2}<\dfrac1{n+1}+\dfrac1{(n+1)^2}+\dfrac1{(n+1)^3}+...=\dfrac{\dfrac1{1+n}}{1-\dfrac1{n+1}}=\dfrac1n$

Instead of comparing $\frac{1}{(n+1)^2} + \frac{1}{n+1}$ and $\frac{1}{n}$, we can compare $\frac{1}{(n+1)^2}$ and $\frac{1}{n}-\frac{1}{n+1}$. Then, what we have is $$\frac{1}{(n+1)^2} = \frac{1}{(n+1)(n+1)} < \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} \implies \frac{1}{(n+1)^2} < \frac{1}{n}-\frac{1}{n+1}$$ $$\implies \frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$$

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Alternatively, by using the same way, we could try to prove $$\frac{1}{n}-\frac{1}{(n+1)^2}-\frac{1}{(n+1)} > 0$$ For this one, we have $$\frac{1}{n}-\frac{1}{(n+1)^2}-\frac{1}{(n+1)} = \frac{(n+1)^2-n-n(n+1)}{n(n+1)^2} = \frac{1}{n(n+1)^2} > 0$$ since $n \ge 1$. Therefore, the result follows.

Multiply through by $n(n+1)^2$ (a positive number since $n \ge 1$ ) to get $$n+n(n+1) < (n+1)^2$$ $$n < (n+1)^2 - n(n+1)$$ $$n < (n+1)(n+1-n)$$ $$n < n+1$$ Which is true of course for all $n$; in particular $n\ge 1$