On the degree elevation needed to bring Bernstein coefficients to [0, 1] Take a polynomial $f(x)$ of even degree $n$ of the form—$${n \choose {n/2}}x^{n/2}(1-x)^{n/2} k,$$where $k>1$ is the $(n/2)$th Bernstein coefficient of the polynomial.  (With these properties, $f$ peaks at the point 1/2 in the interval [0, 1] and is nonnegative everywhere in [0, 1].)
Suppose $f(1/2) \in (0, 1)$, and suppose the polynomial's degree is elevated enough times that all its Bernstein coefficients are in the interval $[0,1]$.  Let $r$ be the smallest number of degree elevations needed before this happens.
Then experiments show that $r/n$ appears to have a limit of $1/3$ as $n$ approaches infinity.
I further conjecture that it is enough to elevate $f(x)$, floor($n/3$)+1 times to bring all its Bernstein coefficients to the interval $[0, 1]$.
My question is:  Is there a proof of these claims?  It is nothing I could find so far in the papers on Bernstein polynomials.

EDIT:
An answer provided a useful lower bound on the number of degree elevations, namely $r \ge \frac{nf(1/2)^2}{1-f(1/2)^2} = m$.  My new question is:  Is it enough to elevate $f(x)$, floor($m$)+1 times to bring all its Bernstein coefficients to the interval $[0, 1]$?
 A: We have $$ f(x) = {n \choose {n/2}}x^{n/2}(1-x)^{n/2} k = 2^n f(1/2) x^{n/2}(1-x)^{n/2} = 2^n f(1/2) x^{n/2} (1-x)^{n/2} ( x+ (1-x))^r = \sum_{j=0}^r 2^n f(1/2)  \binom{r}{j}  x^{n/2+j}(1-x)^{n/2 + r-j} $$
so the $n/2+j$th Bernstein coefficient is $$ \frac{  2^n f(1/2)  \binom{r}{j}  } { \binom{n +r }{ n/2 + j } } .$$
These are certainly nonnegative, so it suffices to check when these are at most $1$. Increasing $j$ by one multiplies this expression by $\frac{r-j}{j+1} \frac{ n/2 +j+1}{ n/2 +r-j} = \frac{ (r-j)(j+1) + (r-j)(n/2) }{ (r-j)(j+1) + (j+1) (n/2)}$ which increases when $j+1< r-j$ and decreases when $j+1< r-j$, so the maximum is attained at $j=r/2$ (or $\frac{r \pm 1}{2}$ if $r$ is odd.)
Thus ($r$ even, for simpilicity), it suffices to know when the inequality
$$ 2^n \binom{r}{r/2} f(1/2) \leq \binom {n+r}{(n+r)/2}$$ holds.
Since $\binom{r}{r/2} \approx 2^r \sqrt{\frac{2}{ \pi r}}$ and $\binom {n+r}{(n+r)/2}\approx 2^{n+r}  \sqrt{\frac{2}{ \pi(n+r}}$, this will happen roughly when $ (n+r) f(1/2)^2 \leq r$ or $r \geq \frac{n f(1/2)^2 }{ (1 - f(1/2)^2 }$.
So I think the answer depends a lot on how much smaller $f(1/2)$ is than $1$...
