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.


If $f(x)=(1-x)e^{2x}$, then $f'(x)=(1-2x)e^{2x}=0$ when $x=\frac{1}{2}$. Drawing a graph/checking the second derivative shows it to be a maximum, whence $1=f(0)\le f(x)\le f(1/2)=\frac{e}{2}$ on $[0,\frac{1}{2}]$. We thus have:
$$1\le(1-x)e^{2x}\le\frac{e}{2}$$
$$\implies \frac{1}{1-x}\le e^{2x}\le\frac{e}{2(1-x)}$$
$$\implies \frac{1}{\sqrt{1-x}}\le e^x \le \sqrt{\frac{e}{2}}\frac{1}{\sqrt{1-x}}$$
on the given interval.
The reason I chose this approach is that much in the same way as young children make most of their arithmetic mistakes when dealing with fractions and negative numbers, I find myself far more at ease when fractions, square roots, inverse functions and such are all cleared out (I'm still very averse to the quotient rule for differentiation). So dealing with $(1-x)e^{2x}$ is greatly preferable for me, and is set up in such a way that the required bounds should pop out quite naturally.
