Inequality for exponential of $x^\alpha$

Question

Let $\alpha\in (0,1)$. Is there an inequality for the function $f(x)=e^{|x|^{\alpha}}$ so that we can get rid of $x^{\alpha}$?. Something like: $$e^{|x|^{\alpha}} \leq (C_1+C_2)e^{C_3|x|},$$ for some constants $C_1,C_2,C_3>0$ depending only on $\alpha$? :) It seems so but I'm not 100% sure.

Actually, thinking of it, any function $|x|^{\alpha}$, $\alpha\in (0,1)$ has linear growth right? i.e. $|x|^{\alpha}\leq C(1+|x|)$ for some constant. So the question would be how $C$ depends on $\alpha$.

Thanks for the help!

Answer 1

$$ |x|^\alpha \le 1 + |x| $$ is almost obvious, just consider the two cases $|x| \le 1$ and $|x| > 1$.

The weighted AM-GM inequality gives $$ |x|^\alpha = 1^{1-\alpha} |x|^\alpha \le (1-\alpha) + \alpha |x| \le \max(\alpha, 1-\alpha) (1 + |x|) $$ for $0 < \alpha < 1$ and arbitrary $x \in \Bbb R$.

The minimum of $x^\alpha/(1+x)$ for $x > 0$ and fixed $\alpha \in (0, 1)$ is $$ C = (1-\alpha)\left(\frac {\alpha}{1-\alpha} \right)^\alpha $$ and attained at $x = \alpha/(1-\alpha)$. This is a slightly better bound:

Answer 2

For $|x| \leq 1$, bound $|x|^\alpha \leq 1$. Otherwise $|x|^\alpha \leq |x|$.