Let $X$ be a random variable following a $\mathrm{Binomial}(n,p)$ distribution, and let $$Y=\min\{X,n-X\}.$$ Ispired by the problem posed by C. Clement on <https://math.stackexchange.com/questions/1696256/expectation-and-concentration-for-minx-n-x-when-x-is-a-binomial>, I want to ask whether there exists some constant $c>0$ such that $\mathbb{E}(Y)\geq c\cdot\min\{p,1-p\}\cdot n$ for all $0<p<1$. If this is not true, can we find some $p\_0$ with $\frac{1+\sqrt{5}}{4}\leq p\_0<1$ such that if $X\sim \mathrm{Binomial}(n,p\_0)$ then $\mathbb{E}(Y)\geq (1-p)[(1+4p)n-8(1+p)]$?



A lower bound for the expectation of $\min\{X,n-X\}$ when $X$ follows a $\mathrm{Binomial}(n,p)$ distribution
Since $\min(a,b)=(a+b-|a-b|)/2$, your question is really about upper bounding $E|2X-n|$, or, equivalently, $E|X-n/2|$:


$$E\min(X,n-X)=n/2-2E|X-n/2|.$$


You can upper bound $E|X-n/2|$ using Jensen's inequality:
$E|X-n/2|\le\sqrt{E(X-n/2)^2}$.


The latter, if I'm not mistaken, evaluates to
$$ n\sqrt{
p(1-p)/n+p^2-p+1/4
}
=:nF(p).$$


For $n$ sufficiently large and $p$ sufficiently small (certainly, $p\le 0.65$; the exact value can be easily computed), we have
$2F(p)\le c(1-p)$
for some universal $c>0$.
That means
that
$n/2-2F(p)n\ge cnp$, so your conjecture holds for this range of $p$. [I'm confident that with a bit more care you can extend the result to all $p$, perhaps with a worse constant. Update: this "confidence" has proven misplaced, see Dmitry Krachun's answer below!]


