Extension to a different base of an inequality (upper bound) to the hyperbolic cosine

Question

For any $a\in\mathbb{R}$, it is pretty straightforward to show, using the Maclaurin series for $e^a$ and $\cosh(a)$ that

$$ \cosh(a) \le e^{a^2/2} $$

I would like to show a similar bound for $$ \frac{b^a + b^{-a}}{2} $$ for $b>1$, which one could interpret as the hyperbolic cosine with a different basis. I hoped for $$ \frac{b^a + b^{-a}}{2} \le b^{a^2/2} $$ but I cannot show it (which does not mean it cannot be shown, btw).

Is there any other bound available? I know I could get $b^{|a|}$, but since $a$ is likely to be in $(-1,1)$, would prefer something with higher exponent for $a$.

Answer 1

Try $b=4$ and $a=1$. Then $$ \frac{b^a+b^{-a}}2=\frac{17}8\gt2=b^{a^2/2}\tag{1} $$ which shows that $\frac{b^a+b^{-a}}2\le b^{a^2/2}$ is not true for all $b\gt1$.

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If we know that $\frac{e^a+e^{-a}}2\le e^{a^2/2}$, then $$ \begin{align} \frac{b^a+b^{-a}}2 &=\frac{e^{a\log(b)}+e^{-a\log(b)}}2\\ &\le e^{(a\log(b))^2/2}\\[5pt] &=b^{\log(b)a^2/2}\tag{2} \end{align} $$ and if $1\le b\le e$, then $(2)$ implies $$ \begin{align} \frac{b^a+b^{-a}}2 &\le b^{\log(b)a^2/2}\\ &\le b^{a^2/2}\tag{3} \end{align} $$

Answer 2

Hint:

$$b^a=e^{a\log b}$$

and

$$\cosh(x)=\frac{e^x+e^{-x}}{2}$$