TITLE: Prove that $f(x,y)+f(y,z)\ge f(x,z)$ where $f(x,y)=\sqrt{x\ln x+y\ln y-(x+y)\ln(\frac{x+y}2)}$
QUESTION [7 upvotes]: Denote $f(x,y)=\sqrt{x\ln x+y\ln y-(x+y)\ln(\frac{x+y}2)}$.
Show that $f(x,y)+f(y,z)\ge f(x,z)$ for $x,y,z> 0$.

This is a question from a friend, which is a deep learning homework. It looks like some square roots of KL divergence, but it seems no help. Some other friends have tried to square it but dealing with the crossing terms like $xy\ln x \ln y$ makes it tough. Some friends and the asker himself try to take the derivative of $y$ and calculate the minimum but to no avail... Some students suggested that it can be written as an integral. I don't have any idea, so I ask here.

REPLY [5 votes]: Complement to @Carl Schildkraut's very nice answer:
Remarks:

Alternatively, we just prove that, for all $x, y > 0$,
$$\int_0^\infty \left(\frac{\mathrm{e}^{-t x/2} - \mathrm{e}^{-ty/2}}{t}\right)^2\mathrm{d} t = x\ln x + y\ln y - (x + y)\ln \frac{x + y}{2}.$$
Then apply Minkowski's integral inequality.

Alternatively, using the identity $\frac{1}{u + v} = \int_0^1 t^{u + v - 1} \mathrm{d} t$ (rather than $\frac{1}{u + v} = \int_0^\infty \mathrm{e}^{-t(u + v)}\mathrm{d} t$), we may turn to prove that
$$\int_0^1 \left(\frac{t^x - t^y}{\sqrt{2t}\ln t}\right)^2 \mathrm{d} t = x\ln x + y\ln y - (x + y)\ln \frac{x + y}{2}.$$
Then apply Minkowski's integral inequality.


$\phantom{2}$

We have
$$\int_{x/2}^{y/2} \int_{x/2}^{y/2}
\frac{1}{u + v}\mathrm{d}u\mathrm{d} v = x\ln x + y\ln y - (x + y)\ln\frac{x + y}{2}.$$
Also, we have, for all $u, v > 0$,
$$\int_0^\infty \mathrm{e}^{-t(u + v)} \mathrm{d} t
= \frac{1}{u + v}.$$
We have
\begin{align*}
 f(x, y) &= \int_{x/2}^{y/2} \int_{x/2}^{y/2}
 \frac{1}{u + v}\mathrm{d}u\mathrm{d} v \\
 &= \int_{x/2}^{y/2} \int_{x/2}^{y/2}
 \int_0^\infty \mathrm{e}^{-t(u + v)} \mathrm{d} t \mathrm{d}u\mathrm{d} v\\
 &= \int_0^\infty \left(\int_{x/2}^{y/2} \mathrm{e}^{-tu}\mathrm{d}u\right) \left(\int_{x/2}^{y/2} \mathrm{e}^{-tv}\mathrm{d}v\right) \mathrm{d} t \\
 &= \int_0^\infty \left(\frac{\mathrm{e}^{-t x/2} - \mathrm{e}^{-ty/2}}{t}\right)^2\mathrm{d} t.
\end{align*}
It suffices to prove that
\begin{align*}
&\sqrt{\int_0^\infty \left(\frac{\mathrm{e}^{-t x/2} - \mathrm{e}^{-ty/2}}{t}\right)^2\mathrm{d} t} + \sqrt{\int_0^\infty \left(\frac{\mathrm{e}^{-t y/2} - \mathrm{e}^{-tz/2}}{t}\right)^2\mathrm{d} t}\\
\ge\, & \sqrt{\int_0^\infty \left(\frac{\mathrm{e}^{-t x/2} - \mathrm{e}^{-tz/2}}{t}\right)^2\mathrm{d} t}
\end{align*}
which is true, using Minkowski's integral inequality ($p > 1$):
$$\left(\int_a^b |f(x) + g(x)|^p \mathrm{d} x \right)^{1/p}
\le \left(\int_a^b |f(x)|^p \mathrm{d} x\right)^{1/p}
+ \left(\int_a^b |g(x)|^p \mathrm{d}x\right)^{1/p}.$$
We are done.