I have stumbled upon a rather intriguing inequality involving the product of the scaled distribution and the scaled density of a random variable. The inequality has a very attractive form, and it resembles, somehow, Cauchy–Schwarz inequality. The problem goes as follows.

### The Inequality

Let $X$ be a continuous and non-negative random variable. Denote by $\bar F(x)= \mathbb P \{X>x\}$ the complementary cumulative distribution function (ccdf), and $f(x)$ its probability density function (pdf). Given two scaling factors $a_1,a_2\ge1$; consider the *scaled* version of the density function $f_i(\cdot) = f(a_i \cdot)$, and the complementary distribution $\bar F_i(\cdot) = \bar F(a_i \cdot)$.

We need to show that

$$ \langle f_1, \bar F_1 \rangle \langle f_2, \bar F_2 \rangle \ge \langle f_1, \bar F_2 \rangle \langle f_2, \bar F_1 \rangle, $$

where $\langle u, v \rangle = \int_0^\infty u(x)v(x)w(x)\,dx$ is the inner product with weight $w(\cdot)\ge0$. We can assume that all functions are square integrable w.r.t. to the weight.

Has anybody problem seen such an inequality? It has quite an appealing form, but I could neither prove it nor find a counterexample. Any reference or ideas would be appreciated!!

### Remarks

The latter holds when the hazard rate of the random variable $X$ is homogeneous, that is, the hazard rate $h(x) = f(x) / \bar F(x)$ satisfies $h(a x) = a^n h(x)$ for some $n$. In this case, it suffices to write $f_i(\cdot) = a_i^n h(\cdot) \bar F_i(\cdot)$ and use Cauchy–Schwarz. The exponential, weibull, and pareto distributions have homogeneous hazard rates.

It tried numerically with other distributions and it seems to hold.

### Edit: Counter-example

Anthony Quas has provided an excellent counter-example (see his answer below) for general weights. Actually, I was a looking for a particular weight function, which is given by $$w(x) = x \exp(-c_1 \bar F_1(x) - c_2 \bar F_2(x)).$$ Since, the inequality holds for the case of homogeneous hazard rates, I was hoping that it will hold in full generality. Shame on me! Anyway, hope that somebody has some thoughts on this. It would be much appreciated.