Show that $(1 - x)^{-k} \le 1 + 2 k x$ when $0 < k x <1/2$ and $k \ge 1$. I think it is true that
$$(1 - x)^{-k} \le 1 + 2 k x$$ 
when $0 < k x <1/2$ and $k \ge 1$.
This can be proved by using mean-value theorem to show that
$$
(1-x)^{-k} \le 1 + k x / (1-x)^{k+1}
\le
1+ 2kx
$$
under the given condition.
Is there's a more direct way to show it?
 A: We have that $0 <  x <1/2$ and by Bernoulli inequality 
$$(1 - x)^{-k} =\frac1{(1 - x)^{k}}\le \frac1{1 - kx}$$
and
$$\frac1{1 - kx}\le 1 + 2 k x \iff kx(1-2kx) \ge 0$$
which is true.
A: $(1 - x)^{-k} \le 1 + 2 k x
$
if
$0 \lt x < \dfrac1{2k}$.
This is
$1 
\le (1-x)^k(1+2kx)
$.
If $k=1$ this is
$1
\le (1-x)(1+2x)
=1+x-2x^2
$
or $x \ge 2x^2
$
or
$x \le \frac12$
which is true.
Suppose true for $k$.
Want to show
$1 
\le (1-x)^{k+1}(1+2(k+1)x)
$.
Assume
$0 \lt x < \dfrac1{2(k+1)}$.
$\begin{array}\\
(1-x)^{k+1}(1+2(k+1)x)
&=(1-x)^{k+1}(1+2kx+2x)\\
&=(1-x)^{k+1}(1+2kx)+(1-x)^{k+1}(2x)\\
&=(1-x)(1-x)^{k}(1+2kx)+(1-x)^{k+1}(2x)\\
&\ge(1-x)+\dfrac{(1-x)}{1+2kx}(2x)\\
&=\dfrac{(1+2kx)(1-x)+2x(1-x)}{1+2kx}\\
&=\dfrac{1+(2k-1)x-2kx^2+2x-2x^2}{1+2kx}\\
&=\dfrac{1+(2k+1)x-2(k+1)x^2}{1+2kx}\\
&=\dfrac{1+2kx+x-2(k+1)x^2}{1+2kx}\\
&=1+\dfrac{x-2(k+1)x^2}{1+2kx}\\
&=1+x\dfrac{1-2(k+1)x}{1+2kx}\\
&\ge 1
\qquad\text{if } 1-2(k+1)x \ge 0\\
\end{array}
$
or
$x \le \dfrac1{2(k+1)}$,
which is the
induction hypothesis.
