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

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.

Answer 1

Let $g:\mathbb R_{>0}^2\to\mathbb R_{>0}$ be any function satisfying $$g(s,t)g(u,v)\geq g(s,v)g(t,u)$$ if $u\geq s$ and $v\geq t$. For any region $R\subset\mathbb R_{>0}^2$, define $$\mu(R)=\int_R g(u,v)dudv.$$ Then, for any $0<x\leq y\leq z$, $$\mu\big([x,y]^2\big)\mu\big([y,z]^2\big)\geq \mu\big([x,y]\times [y,z]\big)^2,$$ since \begin{align*} \mu\big([x,y]^2\big)\mu\big([y,z]^2\big) &=\int_x^y\int_x^y\int_y^z\int_y^zg(s,t)g(u,v)\ du\ dv\ ds\ dt\\ &\geq \int_x^y\int_x^y\int_y^z\int_y^zg(s,v)g(u,t)\ du\ dv\ ds\ dt\\ &=\mu\big([x,y]\times [y,z]\big)^2, \end{align*} where we have used that $u\geq s$ and $v\geq t$ everywhere in the region of integration. Define $h(x,y)=\sqrt{\mu\big([x,y]^2\big)}$. Then, for $x\leq y\leq z$, \begin{align*} \big(h(x,y)+h(y,z)\big)^2 &=\mu\big([x,y]^2\big)+\mu\big([y,z]^2\big)+2\sqrt{\mu\big([x,y]^2\big)\mu\big([y,z]^2\big)}\\ &\geq \mu\big([x,y]^2\big)+\mu\big([y,z]^2\big)+2\mu\big([x,y]\times [y,z]\big)\\ &=\mu\big([x,z]\big)^2=h(x,z)^2. \end{align*} This means $h(x,y)+h(y,z)\geq h(x,z)$ for $x\leq y\leq z$. When $z\geq y\geq x$, the inequality is the same, since $h$ is symmetric, and when $y$ is not between $x$ and $z$ one of the terms on the left side exceeds the term on the right. This means that $h(x,y)+h(y,z)\geq h(x,z)$ always.

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Now, define $g(u,v)=\frac1{u+v}$. We have \begin{align*} g(s,t)g(u,v)-g(s,v)g(t,u) &=\frac{(s+v)(t+u)-(s+t)(u+v)}{(s+t)(u+v)(s+v)(t+u)}\\ &=\frac{(u-s)(v-t)}{(s+t)(u+v)(s+v)(t+u)}\geq 0 \end{align*} if $u\geq s$ and $v\geq t$, and \begin{align*} h(x,y) &=\sqrt{\int_x^y\int_x^y\frac1{u+v}du\ dv}\\ &=\sqrt{2x\ln(2x)+2y\ln(2y)-2(x+y)\ln(x+y)}=f(2x,2y). \end{align*} So, $f(x,y)+f(y,z)\geq f(x,z)$ for all $x,y,z>0$, as desired.

Answer 2

Supplement to @Carl Schildkraut's very nice answer:

Remarks:

1. 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.

2. 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.

$\phantom{2}$

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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.

Answer 3

Some hint for another proof :

if we have for $a,b,c,x,p>0$ such that $\exists p\in(0,1/100)$ $$F(x)=g(x,abx/c)-g(x,xa)+g(abx/c,xa)$$

Then we have $x>0$ :

$$F''(x)\leq 0$$

Remark that all the function are concave with a plus so it remains to compare the only positive function in the second derivative of $F$ with another including a squared derivative.

Where :

$$g^2(x,y)=x\ln x+y\ln y+p-(x+y)\ln\frac{x+y}{2}$$ To conclude we can use the minimum of $F(x)$ and the three chord lemma as $F(0)=0$

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