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.


For our interval, the inequality is equivalent to $1-x\ge e^{-2x}$. (We squared and flipped.)
This inequality can be proved using differential calculus. Let $f(x)=1-x-e^{-2x}$. Then $f'(x)=2e^{-2x}-1$. So $f(x)$ is increasing until $x=\frac{\ln 2}{2}\approx 0.34$ and then decreasing.  Thus all we need to do is check its value at $x=1/2$. 
