# A simple two variable analytic inequality, inspired by probability

I'm trying to prove the following inequality: $$bf_1g_1 + (x-b)f_1g_0 + (y-b)f_0g_1 + (1-x-y+b)g_0f_0 \le (|f_1|^p x + |f_0|^p (1-x))^{1/p} (|g_1|^p y + |g_0|^p (1-y))^{1/p}$$ where $0\le xy\le b\le y\le x\le 1-y$, $f_1,f_0,g_1,g_0\in\mathbb{R}$ and $p=\log\left(\frac{(1-x)(1-y)}{xy}\right){\large/}\log\left(\frac{1-x-y+b}{b}\right)$.

Another way to write this is $\langle F,G\rangle\le\|F\|_p\|G\|_p$ where $F$ and $G$ are two-value random variables over $\{f_1,f_0\}$ and $\{g_1,g_0\}$ respectively, and $\begin{pmatrix}b&x-b\\y-b&1-x-y+b\end{pmatrix}$ is their joint probability matrix.

I can reduce to the following: $$b(xy)^{-1/p}(1+f)(1+g) + (x-b)(x(1-y))^{-1/p}(1+f)(1-g) + (y-b)((1-x)y)^{-1/p}(1-f)(1+g) + (1-x-y+b)((1-x) (1-y))^{-1/p}(1-f)(1-g) \le ((1+f)^p + (1-f)^p)^{1/p} ((1+g)^p + (1-g)^p)^{1/p}$$ where $-1\le, f, g\le 1$.

I can identify two points where the curves $LHS(f,g)$ and $RHS(f,g)$ are equal and in which their derivatives in $f$ and $g$ are equal as well. I conjecture that these are the only such points backed up by the contour plot of $RHS-LHS$ below, plotted for $b=.1$, $y=.2$ and $x=.3$:

I was hoping to use the fact that $RHS$ is convex and $LHS$ is linear in $f$ and $g$ to conclude the inequality for the entire range of $f,g$, however, the combined function $LHS(f,g)$ is clearly not linear.

If anyone has suggestions for how I may proceed I'd be very grateful.

The inequality in question is false when e.g. $f_0=f_1=g_0=1$, $g_1=2$, $x=1/2$, $y=1/2-h$, $b=h$, and $h>0$ is small enough; for instance, $h=1/20$ will do here to disprove the inequality.
• Sorry, I should add $b>xy$. For $b=xy$ we get equivalence (easy to check) and for $b<xy$ we should get the inverse inequality. – Thomas Dybdahl Ahle Aug 9 '18 at 17:01