$\int_0^1 xf(x)\,dx \le \int_0^1 \frac{2}{3}f(x)\,dx$ for a concave function $f(x)$

Question

Let $f$ be a concave function (therefore continuous) on $[0,1]$. I want to prove that $\int_0^1 xf(x)\,dx \le \int_0^1 \frac{2}{3}f(x)\,dx$ and find the condition when $=$ holds.

I have tried using Riemann Sums, LHS is the limit of $\sum_i \frac{i}{n^2} f(\frac{i}{n})$ and RHS is the limit of $\sum_i \frac{2i}{3n} f(\frac{i}{n})$, but I don't know how to continue (this may not work).

Answer 1

HINT:

Assume $f(0)\ge 0$.

First, integrate by parts with $u=x$ and $v=\int_0^x f(t)\,dt$.

Then, use $\int_0^x f(t)\,dt\ge \frac12 xf(x)$ for any concave $f(x)$.

Answer 2

This is false as it stands, $f(x)\equiv -1$ is a counterexample.