Prove that $\int ^1_0x^2f(x)dx\cdot\int ^1_0f(x)dx\leq \int ^1_0xf(x)dx\cdot\int^1_0f(x)^2dx$

Question

Let $f:[0,1]\to (0,+\infty)$ be a decreasing function. Prove that:

$$\int ^1_0x^2f(x)dx\cdot\int ^1_0f(x)dx\leq \int ^1_0xf(x)dx\cdot\int^1_0f(x)^2dx$$

My approach: I think it can be solved by C-S inequality. I tried to make those expressions appear as below: $$\int ^1_0f(x)^2dx\cdot\int ^1_01dx\geq \big(\int ^1_0f(x)dx\big)^2$$ and
$$\int ^1_0xf(x)dx\geq \int ^1_0x^2f(x)dx, \forall x\in [0,1]$$
But it doesn't lead to the desired inequality. I know I haven't used the condition about decreasing function yet but I don't know how to handle it.

Answer 1

Are you sure this is correct?

Let's assume initially that you do not mean f is strictly decreasing. Then I can take $f(x)\equiv c$ for some $c>0$.

Then the inequality to prove becomes $\int_0^1 cx^2 dx\cdot\int_0^1 cdx\leq \int_0^1 cx dx\cdot\int_0^1 c^2 dx$, or equivalently $c^2\frac{1}{3}\leq c^3\frac{1}{2}$ and $c\geq \frac{2}{3}$. In particular if I take $f\equiv\frac{1}{2}$, the inequality will fail.

Now if you want a counterexample that is strictly decreasing, using the continuity of the integral operator you can take any very mildly decreasing function that starts with $f(0)=\frac{1}{2}$, say something like for a small $\varepsilon>0$ the linear function $f(x)=\frac{1}{2}-\varepsilon x$.

If you plug this function into the inequality you will get the constant case with $c=\frac{1}{2}$ and a bunch of other terms multiplied by powers of $\varepsilon$. Choosing $\varepsilon$ sufficiently small will therefore violate the inequality.