How to prove inequality? What is an easy way to show that for positive integers $i,n$, a real $p \in (\frac12,1)$ and $\epsilon \in [0,p]$,
$$p^i(1-p)^{n-i} \geq (p-\epsilon)^i(1-(p-\epsilon))^{n-i}.$$
(I have a complicated way, where I first show that the left hand side is bigger when i = n/2 and then increasing i can only make the left hand side bigger. But is there some well known inequality which lets me formulate this shorter?)
 A: This inequality cannot be true.  You need more conditions.
If $p$ is allowed to be larger than $1$ there will be sign problems since $n-i$ odd will make both sides negative.  Even if we suppose $0<p<1$ and $0<e<p$ it does not work out.
Consider the case where $0.5n<i<0.6n<n\log_3 ( 2)$.  Let $p=\frac{3}{4}$, and let $e=\frac{1}{4}$.  Then the inequality becomes $$\left(\frac{3}{4}\right)^{i}\left(\frac{1}{4}\right)^{n-i}\geq\left(\frac{1}{2}\right)^{i}\left(\frac{1}{2}\right)^{n-i}$$ or equivalently $$\frac{3^{i}}{2^{n}}\geq1.$$ Since $i<n\log_{3}2$ we have that $3^{i}<2^{n}$ so that $$\frac{3^{i}}{2^{n}}<1$$ which is impossible.
A: It is not true.  If $p=.8, n=5, i=3, \epsilon =.01, p^i(1-p)^{n-i}=.2048, (p-\epsilon)^i(1-(p-\epsilon))^{n-i}\approx .21743$  If you take the derivative $\frac{d(p^i(1-p)^{n-i})}{dp}=p^{i-1}(1-p)^{n-i-1}(i-np)\lt 0 \text { if } i-np \lt 0$, so if you decrease $p$ you increase the function.
A: Let $F: (0,1) \rightarrow R, f(x)= x^i (1-x)^{n-i}$.
Then $f'$ is positive on $(\frac{i}{n}, 1)$ and negative on $(0,\frac{i}{n})$.
Thus your inequality holds if $p-\epsilon > \frac{i}{n}$, and is opposite if $p< \frac{i}{n}$. The left out case is much harder to stude.
