Integral inequality - Proof techniques

I would like to prove that the function $$F(x) = \frac{1}{a^2}\int_0^a f(s) s \mathrm{d}s - \frac{1}{x^2}\int_0^x f(s) s \mathrm{d}s$$ is non-positive for $x \in [0, a]$, given that the function $f$ is continuous, positive and non-increasing over the same interval.

I think I managed as I show afterwards: I still would like to see how it could be done more concisely/elegantly/effectively.

My first approach: firstly I compute the derivative of $F$, $$F ^\prime (x) = \Big[- \frac{1}{x^2} f(x) x + \frac{2}{x^3} \int_0^x f(s) s \mathrm{d}s \Big]$$

As $f(x)$ is non-increasing, the following inequality stands

$$\frac{2}{x} \int_0^xf(s) s \mathrm{d}s \geq 2f(x) x$$ and then $$F ^\prime(x) = [- \frac{1}{x^2} f(x) x + \frac{2}{x^3} \int_0^x f(s) s \mathrm{d}s ] \geq \frac{1}{x^2} [-f(x) x +2f(x) x] = \frac{1}{x^2} f(x) > 0$$

I compute then $F(0)$. As $$\lim_{x\to 0} \frac{1}{x^2} \int_0^x f(s) s \mathrm{d}s = \frac{1}{2}f(0)$$ it follows that $$F(0) = \frac{1}{a^2}\int_0^a f(s) s \mathrm{d}s - \frac{1}{2}f(0)$$ which is non-positive (it is actually nihil when $f$ is constant) As $F(0) \leq 0$, $F(a) = 0$ and $F$ is non-decreasing, I conclude it must be non-positive, a desired.

The second approach I thought about is consists in arguing that the expression $$\frac{x^2} {a^2} \leq \frac { \int_0^x f(s) s \mathrm{d}s } {\int_0^a f(s) s \mathrm{d}s }$$ stands. One notices that the equality stands if $f(x)$ is constant. As $f(x)$ is non increasing, the second term must be larger than in the case of costant $f(x)$, which proves the claim.

As said, I would like to know how the same claim can be proved in more efficient/elegant/coincise fashion.

Note that $F(x) \leq 0$ can be equivalently written as $$\frac{\int_{0}^x f(s)s \,ds}{\int_0^x s\,ds} \geq \frac{\int_{0}^a f(s)s \,ds}{\int_0^a s\,ds}.$$ From Cauchy's mean value theorem, we can show that $\exists \,\xi \in (0, x), \eta \in (x, a)$ such that $$f(\xi) = \frac{\int_{0}^x f(s)s \,ds}{\int_0^x s\,ds}, \,f(\eta) = \frac{\int_{x}^a f(s)s \,ds}{\int_x^a s\,ds}.$$ Since $f(x)$ is non-increasing, we have $f(\xi) \geq f(\eta)$, i.e., $$\frac{\int_{0}^x f(s)s \,ds}{\int_0^x s\,ds} \geq \frac{\int_{x}^a f(s)s \,ds}{\int_x^a s\,ds}.$$ As a result, $$\frac{\int_{0}^x f(s)s \,ds}{\int_0^x s\,ds} \geq \frac{\int_{0}^x f(s)s \,ds + \int_{x}^a f(s)s \,ds}{\int_0^x s\,ds + \int_x^a s\,ds} = \frac{\int_{0}^a f(s)s \,ds}{\int_0^a s\,ds},$$ where we haved use the following fact for ratios：
Suppose $a, b, c, d > 0$, then $$\frac{a}{b} \geq \frac{c}{d} \implies \frac{a}{b} \geq \frac{a + c}{b + d} .$$
In general, when $f(x), g(x)$ are positive and continuous, and $f(x)$ is non-increasing, we have $$\frac{\int_{0}^x f(s)g(s) \,ds}{\int_0^x g(s)\,ds} \geq \frac{\int_{0}^a f(s)g(s) \,ds}{\int_0^a g(s)\,ds}.$$