$0<e^{-\sqrt x}<\frac{1}{x^a}$ inequality

Question

How can we prove that $0<e^{-\sqrt x}<\frac{1}{x^a}$ is true for all $x>B(a)$ for some $B(a)$ and some fixed $a>0$. By the mean theorem I can only receive $\frac{1}{1+\sqrt x}>e^{-\sqrt x}$ and after that $\frac{1}{\sqrt x}>e^{-\sqrt x}$. Then I have next inequality for $x>1$ because $\frac{1}{x^b}>\frac{1}{\sqrt x}>e^{-\sqrt x}$ but it I have only for $b<1/2$. Thanks for your help.

Answer 1

The inequality is equivalent to $-\sqrt{x}<-a\log x$ or $\sqrt{x}>2a\log\sqrt{x}$. Set $t=\sqrt{x}$ and consider the function $$ f(t)=t-2a\log t $$ Since $\lim_{t\to\infty}f(t)=\infty$, there is some number $C(a)$ such that $f(t)>0$ for $t>C(a)$. Now set $B(a)=C(a)^2$.

Answer 2

Rewrite our inequality in the following form: $$\frac{\sqrt{x}}{\ln{x}}>a.$$ Now, prove that $\frac{\sqrt{x}}{\ln{x}}>\ln{x}$ for all $x>5510$ and we can take $$B(a)=\max\left\{5510,e^a\right\}+1$$