Prove this power inequality $ (1 - F + x)^{s-1} \geq \frac{(1 - F + x)^s - (1 - F)^s}{sx}$ I want to show that
$$ (1 - F + x)^{s-1} \geq \frac{(1 - F + x)^s - (1 - F)^s}{sx}$$
for $x \in [0, F]$, $F\in (0, 1]$, and $s$ is a positive integer larger than one.
I can't find any obvious identity to be used here. The only strategy I could come up with was that for $x\to 0$, the right-hand side becomes identical to the left-hand side. So then it's sufficient to show that
$$ \frac{\partial}{\partial x}  (1 - F + x)^{s-1} \geq \frac{\partial}{\partial x}\frac{(1 - F + x)^s - (1 - F)^s}{sx} \qquad \forall x \in (0, F)$$
Which appears to be true numerically. However, the derivatives look even more messy:
$$ (s-1)(1 - F + x)^{s-2} \geq \frac{s(1-F+x)^{s-1} - \frac{s}{x}\left[(1-F+x)^s - (1-F)^s\right]}{sx}.$$
Is there any path forward here?
 A: $\frac{(1 - F + x)^s - (1 - F)^s}{sx} = (\frac{(1-F+x)-(1-F)}{sx})\sum_{k=0}^{s-1}(1-F+x)^k(1-F)^{s-1-k}=\frac{1}{s}\cdot\sum_{k=0}^{s-1}(1-F+x)^k(1-F)^{s-1-k} \leq \frac{1}{s}\cdot\sum_{k=0}^{s-1}(1-F+x)^k(1-F+x)^{s-1-k}=\frac{1}{s}\cdot\sum_{k=0}^{s-1}(1-F+x)^{s-1}=\frac{s(1-F+x)^{s-1}}{s} =(1-F+x)^{s-1} $
Some of the minute details maybe need to be shown, but I'll leave that up to you.
Sorry about the spacing. I haven't been on here in a while.
If it's still not clear, rewrite it with $A = 1-F$ and $n=s-1$.
Hope this helps.
A: Let $C(x)=1-F+x$ and let $H(x)=xsC(x)^{s-1}-C(x)^s+(1-F)^s.$
For $x>0$ and $s>0$ the inequality is equivalent to $H(x)\ge 0.$
We have $H(0)=0.$
For a fixed $F$ and fixed $s\ge 2$ we have $C'(x)=1$ so $$H'(x)=sC(x)^{s-1}+xs(s-1)C(x)^{s-2}-sC(x)^{s-1}=xs(s-1)C(x)^{s-2}$$ which   is positive when $0<x<F\le 1.$
A: Apply the Mean Value Theorem to the function $f(t)=t^s$, with $t_1= 1-F$ and $t_2=1-F +x$,
$$\frac{(1 - F + x)^s - (1 - F)^s}{x} = f'(c)=sc^{s-1} \le s(1 - F + x)^{s-1}$$
where used in the last step are the facts that $t^{s-1}$ of $s>1$ monotonically increases  and $c \le 1 - F + x$. Thus, 
$$ (1 - F + x)^{s-1} \geq \frac{(1 - F + x)^s - (1 - F)^s}{sx}$$
