TITLE: Let $f : [0,1] \to \mathbb{R}$, prove that $2 \int_{0}^{1} f(x)dx \ge f\Big(\frac{1}{n}\Big) + \sum_{k=1}^{n-1}\frac{1}{k} f\Big(\frac{k}{n}\Big)$
QUESTION [5 upvotes]: Let $f : [0,1] \to \mathbb{R}$ be a differentiable function with a continuous derivative such that $f(x) \ge xf'(x), \forall x \in [0,1]$. Prove that:
$$2 \int_{0}^{1} f(x)dx \ge f\Big(\frac{1}{n}\Big) + \sum_{k=1}^{n-1}\frac{1}{k} f\Big(\frac{k}{n}\Big)$$, $\forall n \in \mathbb{N}, n \ge 2$
$\mathbf{Edit}$: Like Elaqqad and Crostul pointed out, there may be a typo in the inequality. What if we have:
$$2 \int_{0}^{1}f(x)dx \ge f\Big(\frac{1}{n}\Big) + \sum_{k=1}^{n-1}\frac{1}{n} f\Big(\frac{k}{n}\Big)  ?$$
Will this inequality stand?

REPLY [3 votes]: The problem here is that your inequality is not true, take for example the constant function: $\forall x\in [0,1]$ $$f(x)=1$$ 
your inequality is equivalent to :
$$2=2 \int_{0}^{1} f(x)dx \ge f\Big(\frac{1}{n}\Big) + \sum_{k=1}^{n-1}\frac{1}{k} f\Big(\frac{k}{n}\Big)=1+H_{n-1}=1+\sum_{k=1}^{n-1}\frac{1}{k}$$
which is clearly false, because:
$$H_n \sim \ln(n)\xrightarrow{n\to \infty} \infty$$
(I don't know how we can correct it, but the difference is very large for almost constant functions)
Edit
The proposed correction does not hold either, take for example $n=2$, 
$$2 \int_{0}^{1} f(x)dx \geq \frac{3}{2} f \left(\frac{1}{2}\right) $$ which does not hold for a function like $f(x)= x(1-x)$.
As a correction the only inequality I could think of is the Reimann inequality holding for the decreasing function $g(x)=\frac{f(x)}{x}$ and I think it's the only one which links between $f\left(\frac{k}{n}\right)$ and the integral of $g$