Approximating the area below average of a concave function Given a non-decreasing concave function $f:[0,1]\rightarrow \mathbb{R}^+$. Define 

\begin{align*}
F(n)=\displaystyle\sum_{i=1}^n \min\left\{\frac{1}{n},\frac{f\left(\frac{i}{n+1}\right)}{n+1}\right\}
\end{align*}

We want to find a tight lower bound for $F(n)$ in terms of $n$, knowing that $\int_0^{1} f(x) dx =1 $. 

For instance, I am able to prove that $F(1)\geq \frac{1}{2}$, $F(2)\geq \frac{2}{3}$, $F(3)\geq \frac{1}{3}+\frac{3}{8}$ and see that they are tight for the function $f(x)=2x$. Also, I can prove that $\displaystyle\lim_{n\rightarrow \infty} F(n) \geq 3/4$, and again observe that it is tight for the function $f(x)=2x$. 
 Thanks for your time and attention!
$f(x)$ is non-decreasing. For given $n$, split the range of the sum at $n^*$,
where
$$
\frac{f(n^* / (n + 1))}{n + 1} = \frac{1}{n}
$$
Then the sum is:
$$
F(n) = \frac{1}{n + 1} \sum_{1 \le i \le n^*} f(i / (n + 1)) + (n - n^*) / n
$$
We want to minimize this, and this means keeping as much in the first range (which collects the small values) as we can. By the relation defining $n^*$ we see that it is enough to make sure that $f(x) &lt; \frac{n + 1}{n}$ to have $n^* = n$ for all $n$. By the condition that $\int_0^1 f(x) dx = 1$, this means $f(x) = 1$, and the sum is just $F(n) = \frac{n}{n + 1}$.
$f(x)$ is non-decreasing. For given $n$, split the range of the sum at $n^*$,
where
$$
\frac{f(n^* / (n + 1))}{n + 1} = \frac{1}{n}
$$
Then the sum is:
$$
F(n) = \frac{1}{n + 1} \sum_{1 \le i \le n^*} f(i / (n + 1)) + (n - n^*) / n
$$
We want to minimize this, and this means keeping as much in the first range (which collects the small values) as we can. By the relation defining $n^*$ we see that it is enough to make sure that $f(x) &lt; \frac{n + 1}{n}$ to have $n^* = n$ for all $n$. By the condition that $\int_0^1 f(x) dx = 1$, this means $f(x) = 1$, and the sum is just $F(n) = \frac{n}{n + 1}$.