Prove that $f(1)-f(1/e)\le \int_0^1 \sqrt{x} f'(x) dx$

Let $f:[0,1]\rightarrow \mathbb{R}$ be a differentiable function such that $$f(x^2)+f(y^2)\le2 f(\sqrt{x y}), \space x,y\ge0 $$

Prove that

$$f(1)-f(1/e)\le \int_0^1 \sqrt{x} f'(x) dx$$

Where should I start from?

Solution 1:

This is more of a comment. You could probably tweak this to get the answer.

Note that $$\int_0^1 \sqrt{x} f'(x)dx = \left.\sqrt{x} f(x) \right\vert_0^1 - \int_0^1\dfrac{f(x)}{2\sqrt{x}}dx = f(1) - \int_0^1 \dfrac{f(x)}{2\sqrt{x}}dx$$ Hence, it is enough to prove that $$f(1/e) \geq \int_0^1 \dfrac{f(x)dx}{2\sqrt{x}}$$ From the given inequality, we have $$f(x^2) \leq f(x) \leq f(0)$$ Hence, $$\int_0^1 \dfrac{f(x)}{2\sqrt{x}}dx \leq f(0) \int_0^1 \dfrac{dx}{2\sqrt{x}} \implies f(0) \geq \int_0^1 \dfrac{f(x)}{2\sqrt{x}}dx$$

You can stretch this a bit more. I have weakened it by using the fact that $f(x) \leq f(0)$, instead of $f(x^2) \leq f(x)$. You might try instead to somehow exploit just the fact that $f(x^2) \leq f(x)$. This seems feasible but now I have to run for a meeting.