Proving $\frac{1}{\sqrt{1-x}} \le e^x$ on $[0,1/2]$. Is there a simple way to prove $$\frac{1}{\sqrt{1-x}} \le e^x$$ on $x \in [0,1/2]$?
Some of my observations from plots, etc.:


*

*Equality is attained at $x=0$ and near $x=0.8$.

*The derivative is positive at $x=0$, and zero just after $x=0.5$. [I don't know how to find this zero analytically.]

*I tried to work with Taylor series. I verified with plots that the following is true on $[0,1/2]$:
$$\frac{1}{\sqrt{1-x}} = 1 + \frac{x}{2} + \frac{3x^2}{8} + \frac{3/4}{(1-\xi)^{5/2}} x^3 \le 1 + \frac{x}{2} + \frac{3}{8} x^2 + \frac{5 \sqrt{2} x^3}{6} \le 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \le e^x,$$
but proving the last inequality is a bit messy.


We have
$$-2 \ln \sqrt{1-x}=-\ln(1-x)= \int_{1-x}^1\frac{dt}{t} \leqslant \frac{x}{1-x}.$$
For $0 \leqslant x \leqslant 1/2$, we have $2(1-x) \geqslant 1$ and 
$$-\ln \sqrt{1-x} < \frac{x}{2(1-x)} \leqslant x.$$
Hence,
$$\frac{1}{\sqrt{1-x}} = \exp[-\ln(\sqrt{1-x})]\leqslant e^x.$$
