What is the easiest way to prove the correctness of this inequality I have the following inequality for some $0<x<0.1$: $$x^{1/10}-(1-(1-x)^{x^{-0.5}}) \geq 0$$
Is there an easy way to prove the correctness of such inequality? Thanks!
 A: The inequality in question holds for all $x\in(0,1)$.
Indeed, using the substitution $x=y^{10}$ for $y\in(0,1)$, rewrite the inequality in question as
$$h(y):=\frac{1}{y^5}\,\ln\left(1-y^{10}\right)-\ln(1-y)\ge0.$$
Let
$$h_1(y):=y^6 h'(y)
=y^6 \left(\frac{1}{1-y}-\frac{10 y^4}{1-y^{10}}\right)
-5 \ln\left(1-y^{10}\right).$$
Then
$$h'_1(y)\frac{\left(1-y^{10}\right)^2}{(1-y) y^5}
=5 y^{18}+9 y^{17}+12 y^{16}+14 y^{15}+15 y^{14}+65 y^{13}+64
   y^{12}+62 y^{11}+59 y^{10}+55 y^9+40 y^8+26 y^7+13 y^6
+y^3((y-5)^2+5)
+21 y^2+13 y+6,$$
which is manifestly $>0$ (for $y\in(0,1)$). So, $h'_1>0$ and hence $h_1$ is increasing. Also, $h_1(0+)=0$. So, $h_1>0$ and hence $h$ is increasing. Also, $h(0+)=0$. So, $h>0$ (on $(0,1)$), as desired. $\quad\Box$
A: The desired inequality is true for all $x \in (0, 1)$.
Using Bernoulli's inequality in the form $(1+u)^r\ge 1+ru$ for $r \ge 1$, $u\ge -1$, we have
$$(1 - x)^{x^{-0.5}}
\ge 1 - x \, x^{-0.5} = 1 - x^{1/2}.$$
Thus, we have
$$x^{1/10}-(1-(1-x)^{x^{-0.5}})
\ge x^{1/10} - x^{1/2} \ge 0.$$
We are done.
