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

In an induction proof for $\sum_{k = 1}^n \frac{1}{k^2} \leq 2 - \frac{1}{n}$ (for $n \geq 1$), I was required to prove the inequality $\frac{1}{(n+1)^2} + \frac{1}{n+1} < \frac{1}{n}$.

This is my attempt:

$$ \begin{align} \frac{1}{(n+1)^2} + \frac{1}{n+1} &= \frac{1+(n+1)}{(n+1)^2} \\
&=\frac{n+2}{(n+1)^2} \\
&=\frac{n+2}{n^2 + 2n + 1} \\
&<\frac{n+2}{n^2 + 2n} (\text{since } n \geq 1)\\

&=\frac{n+2}{n(n+2)}\\
&=\frac{1}{n} \end{align} $$

I am just wondering if there is a simpler way of doing this.

Answer

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.