Is $1-\alpha x\le (1-x)^{\alpha}\le 1-\alpha x+\frac{\alpha(\alpha-1)}{2}x^2,$ an inequality about generalized binomial coeffients true?

Is the following inequality true? $$1-\alpha x\le (1-x)^{\alpha}\le 1-\alpha x+\frac{\alpha(\alpha-1)}{2}x^2$$ for real numbers $x,\alpha.$ We may assume $0\le x\le 1$ and put some requirements about $\alpha.$ And $(1-x)^\alpha=\sum_{k=0}^\infty\binom{\alpha}{k}x^k,$ where $\binom{\alpha}{k}=\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}.$
I think I might meet with this inequality in Stanley's book Enumerative Combinatorics 1 http://www-math.mit.edu/~rstan/ec/ec1.pdf or somewhere else, saying that $(1-x)^\alpha$ greater the sum of its expansion terms up to a negative sign, and less than sum up to a positive sign. But I cannot remember it now and I am not sure. And it seems natural if we view it as a Taylor expansion. Any proof, comments or reference are very welcome!

• Have you looked at Bernoulli's inequality? – Jacky Chong Apr 26 '17 at 2:30
• It seems Bernoulli's inequality en.m.wikipedia.org/wiki/Bernoulli%27s_inequality gives the lower bound. How about the upper bound? – Connor Apr 26 '17 at 4:31
• If $\alpha \in (0,1)$ and $x\in (0,1)$ then the left inequality is a strict inequality in the OPPOSITE direction. For example if $\alpha =1/2$ and $x\in (0,1)$ then $1-x/2>0$ so $1-x/2>(1-x)^{1/2}\iff (1-x/2)^2>1-x\iff 1-x+x^2/4>1-x.$ – DanielWainfleet Apr 26 '17 at 9:22

For the upper bound, use \begin{align} (1-x)^r\leq e^{-rx}= 1-rx+ \frac{r^2x^2}{2!}-\frac{r^3x^3}{3!}+\ldots \end{align} for all $x<1$ and $r>0$.
Next, consider the following fact: for $x<0$, we have \begin{align} e^x \leq T_{2n}(x) \end{align} for all $n$ where \begin{align} T_n(x) = 1+x+\frac{x}{2!}+\ldots +\frac{x^n}{n!}. \end{align}
Combining both result yields the desired inequality for all $r>0$ and $0<x<1$.