Prove $\frac{a_n}{S_n^2} \leq \frac{1}{S_{n-1}}-\frac{1}{S_n}$ for partials sums of a divergent series

Some manipulation of the inequality

$$\frac{a_n}{S_n^2} \leq \frac{1}{S_{n-1}} - \frac{1}{S_n} = \frac{S_n - S_{n - 1}}{S_{n-1}S_n} = \frac{a_n}{S_{n-1}S_n}$$

Since $a_n \geq 0$ this is equivalent to $$S_{n-1}S_n \leq S_n^2$$

Since $S_n \gt 0$ this is equivalent to $$S_{n-1} \leq S_n$$

This is true since $S_n - S_{n-1} = a_n \geq 0$.

A start: Note that $$a_n=S_n-S_{n-1}.\tag{1}$$ That holds because $$S_n=(a_1+\cdots+a_{n-1})+a_n=S_{n-1}+a_n.$$ The equality (1) and some manipulation is all you will need. Bring your right-hand side to a common denominator.

You can prove this result by induction. First, in the case $n=2$, \begin{align*} \frac{a_2}{S^2_{2}} \leq \frac{1}{S_{1}}-\frac{1}{S_{2}}, \end{align*} which follows that $\frac{1}{S_{1}}-\frac{1}{S_{2}}= \frac{S_2-S_1}{S_1S_{2}}=\frac{a_2}{S_1S_{2}}$ and $S_2 \geq S_1$.

Suppose that \begin{align*} \frac{a_n}{S^2_{n}} \leq \frac{1}{S_{n-1}}-\frac{1}{S_{n}}. \end{align*} Then \begin{align*} \frac{a_{n+1}}{S^2_{n+1}} = \frac{a_{n+1}}{(S_{n}+a_{n+1})^2} \leq \frac{a_{n+1}}{S_{n}S_{n+1}}=\frac{1}{S_{n}}-\frac{1}{S_{n+1}}. \end{align*} The proof is completed.