Bounds for generalized mean $\frac{1}{2 \mu(a,b)}=\int_0^\infty \frac{dx}{e^{ax}+e^{bx}}$ Let us define a generalized mean of two positive real numbers $\mu(a,b)$ as:
$$\frac{1}{2 \mu(a,b)}=\int_0^\infty \frac{dx}{e^{ax}+e^{bx}}=\int_0^1 \frac{dt}{t^{1-a}+t^{1-b}}$$
This integral has a closed form in terms of hypergeometric function or generalized harmonic numbers (note that we can exchange $a$ and $b$ due to symmetry):

$$\frac{1}{2 \mu(a,b)}=\frac{1}{a} {_2 F_1 } \left(1, \frac{a}{a-b};1+\frac{a}{a-b};-1  \right)$$
$$\frac{1}{2 \mu(a,b)}=\frac{1}{2(a-b)} \left( H_{b/(2(a-b))}-H_{(2b-a)/(2(a-b))} \right)$$

Or, using the hypergeometric series, we actually obtain a very simple form for the integral (assuming $a>b$):

$$\frac{1}{2 \mu(a,b)}=\sum_{n=0}^\infty \frac{(-1)^n}{a+(a-b)n}$$

While I'm mainly interested in finding an interated means algorithm to compute the integral in question (like arithmetic-geometric mean for elliptic integrals), I have started with some bounds.
From experiments we have from below:
$$\mu(a,b) \geq \sqrt{\frac{a^2+b^2}{2}} \geq \frac{a+b}{2}$$

And from above:
$$\mu(a,b) \leq \sqrt[3]{\frac{a^3+b^3}{2}} \leq \frac{a^2+b^2}{a+b}$$

The upper bounds are very sharp for $a$ and $b$ close to each other.
Iterating power means for $p=2,3$ we get a mean which intersects $\mu(a,b)$ (red points).

I tried also intermediate power means $\sqrt[p]{\frac{1+x^p}{2}}$ ($p \in (2,3)$), but they mostly intersect $\mu(1,x)$ close to $x=0$.
However, by solving:
$$2 \mu(1,0)=2^{1/p}$$
We find the better bound:
$$\mu(a,b) \geq \sqrt[p]{\frac{a^p+b^p}{2}}$$
Where:
$$p=\frac{\log (2)}{\log (\log (4))}=2.122089644\dots$$


My questions are:

How can we prove that $\mu(a,b) \leq \sqrt[3]{\frac{a^3+b^3}{2}}$?
Can we prove some sharp bound from below as well? Or a sharper bound from above? (In terms of well known means, or at least elementary functions).

 A: This doesn't derive an iterative method to compute the integral, but it does get a different form in terms of the extended Harmonic numbers (although the value is indeed equal).
Preliminary
$$
\begin{align}
\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k+x}
&=\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac1{k+x}-2\sum_{k=1}^n\frac1{2k+x}\right)\\
&=\lim_{n\to\infty}\left(\sum_{k=n+1}^{2n}\frac1{k+x}+\sum_{k=1}^n\left(\frac1{k+x}-\frac1{k+x/2}\right)\right)\\
&=\log(2)+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x/2}\right)-\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)\\[9pt]
&=\log(2)+H_{x/2}-H_x\\[15pt]
%&=\log(2)+\psi\!\left(\tfrac x2+1\right)-\psi(x+1)
\end{align}
$$
where the Extended Harmonic Numbers are related to the digamma function by $H_x=\gamma+\psi(x+1)$

Integral
Assuming $a\gt b$:
$$
\begin{align}
\int_0^\infty\frac{\mathrm{d}x}{e^{ax}+e^{bx}}
&=\int_0^\infty\frac{e^{-bx}\,\mathrm{d}x}{e^{(a-b)x}+1}\\
&=\sum_{k=1}^\infty(-1)^{k-1}\int_0^\infty e^{-(b+k(a-b))x}\mathrm{d}x\\
&=\frac1{a-b}\sum_{k=1}^\infty\frac{(-1)^{k-1}}{\frac{b}{a-b}+k}\\
&=\left\{\begin{array}{}
\frac1{a-b}\left(\log(2)+H_{\frac b{2(a-b)}}-H_{\frac b{a-b}}\right)&\text{if $a\gt b$}\\
\frac1{2a}&\text{if $a=b$}
\end{array}\right.
\end{align}
$$

An Identity
$$
\begin{align}
H_x+H_{x-1/2}
&=\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x}\right)+\sum_{k=1}^\infty\left(\frac1k-\frac1{k+x-1/2}\right)\\
&=2\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1k-\frac1{2k+2x}-\frac1{2k-1+2x}\right)\\
&=2\lim_{n\to\infty}\sum_{k=1}^{2n}\left(\frac1k-\frac1{k+2x}\right)-2\lim_{n\to\infty}\sum_{k=n+1}^{2n}\frac1k\\
&=2H_{2x}-2\log(2)
\end{align}
$$
This shows that my answer and the value in the question are the same.
A: A few lesser inequalities we can prove directly:
$$\frac{1}{ \mu(a,b)}=\int_0^\infty \frac{2dx}{e^{ax}+e^{bx}} \leq \int_0^\infty e^{-\frac{a+b}{2} x} ~dx$$
$$ \mu(a,b) \geq \frac{a+b}{2}$$
Here we used the AM-GM inequality.
$$\frac{1}{ \mu(a,b)}=\int_0^\infty \frac{2 e^{-ax} e^{-b x} dx}{e^{-ax}+e^{-bx}} \leq \frac{1}{2} \int_0^\infty (e^{-a x} +e^{-b x})~dx=\frac{1}{2} \left(\frac{1}{a} +\frac{1}{b}\right)$$
$$ \mu(a,b) \geq \frac{2ab}{a+b}$$
Here we used the AM-HM inequality.
Obviously, we didn't need the latter bound since we have a better one already, but I wanted to show the principle.
There's a mean, which could allow us to get an upper bound:
$$\frac{a+b}{2} \leq \frac{a^2+b^2}{a+b}$$
In our case, we have:
$$\frac{2}{e^{ax}+e^{bx}} \geq \frac{e^{ax}+e^{bx}}{e^{2ax}+e^{2bx}}$$
$$\frac{1}{ \mu(a,b)} \geq \int_0^\infty \frac{e^{ax}dx}{e^{2ax}+e^{2bx}}+\int_0^\infty \frac{e^{bx}dx}{e^{2ax}+e^{2bx}}$$
$$e^{-ax}=t \\ x=-\frac{1}{a} \ln t \\ dx=-\frac{dt}{at}$$
$$\frac{1}{ \mu(a,b)} \geq \frac{1}{a} \int_0^1 \frac{dt}{1+t^{2(1-b/a)}}+\frac{1}{b}\int_0^1 \frac{dt}{1+t^{2(1-a/b)}}$$
But these integrals are not elementary, they are also expressed in terms of special functions, so it doesn't make much sense to use this bound.
We can try other bounds, but so far I don't see any convenient ones.
