Bernstein polynomials $B_n$ of $f$ where $f$ is concave in $[0,1]$ are nondecreasing with $n$ Problem:
We first try with $f(x)=\sqrt{x}$ for $ x \in [0,1]$ which is clearly concave. 
Defining $B_nf(x)=\sum_{0 \leq k \leq n}\binom{n}{k}f(\frac k n ) x^k (1-x)^{n-k}=\sum_{0 \leq k \leq n}\binom{n}{k}\sqrt{\frac k n } x^k (1-x)^{n-k}$ I would like to prove that if $n \leq m$ then $B_nf \leq B_m f$ i.e. the sequence of Bernstein polynomials of $f$ is nondecreasing.
It is known by Weierstrass theorem that $B_nf$ converges to $f$ uniformly in $[0,1]$ for every $f \in C^0([0,1],\mathbb{R})$.
I do not know how to start because through induction and something similar I cannot reach any conclusion.
Update:
If the result holds in general for $f$ concave, how it can be proven?
 A: Let $f$ be concave on $[0,1].$ Multiplying the definition $$B_nf(x)=\sum^n_{k=0}\binom{n}{k}f\left(\frac k n \right) x^k (1-x)^{n-k}$$ by $1=x+(1-x),$ we have
$$B_nf(x)=\sum^n_{k=0}\binom{n}{k}f\left(\frac k n \right) x^{k+1} (1-x)^{n-k}+\sum^n_{k=0}\binom{n}{k}f\left(\frac k n \right) x^k (1-x)^{n+1-k}.$$ Transforming the summation index in the first sum and adding zero first/last summands, we arrive at
$$B_nf(x)=\sum^{n+1}_{k=0}\left[\binom{n}{k-1}f\left(\frac {k-1} n \right)+\binom{n}{k}f\left(\frac k n \right)\right] x^k (1-x)^{n+1-k},$$ since $\binom{n}{-1}=\binom{n}{n+1}=0.$ Now we have $$\binom{n}{k-1}=\frac{k}{n+1}\binom{n+1}{k}$$ and $$\binom{n}{k}=\left(1-\frac{k}{n+1}\right)\binom{n+1}{k},$$ moreover,
$$\frac{k}{n+1}\cdot\frac {k-1} n+\left(1-\frac{k}{n+1}\right)\cdot\frac k n=\frac{k}{n+1},$$ so we have
$$\binom{n}{k-1}f\left(\frac {k-1} n \right)+\binom{n}{k}f\left(\frac k n \right)\le \binom{n+1}{k}f\left(\frac k {n+1}\right)$$ by concavity of $f,$ and thus
$$B_nf(x)\le B_{n+1}f(x).$$
