Prove that $f(x,y) = \frac{h(xy)}{yh(x) + xh(y)} < 1$

Question

I want to prove that $f(x,y) = \frac{h(xy)}{yh(x) + xh(y)} < 1$ in $(x,y) \in (0,1)^2$, where $h(x) = -x\log_2x - (1-x)\log_2(1-x)$ is the binary entropy function. Proving that it is bounded by some arbitrary constant $C$ instead of $1$ would also suffice. I've tried directly calculating the value of $f$ after plugging $h$ and applying some bounds from the wiki page of $h$, but to no success.

Answer 1

With $h(0) = h(1) = 0$ the function $h$ becomes continuous on $[0, 1]$. Then $$ F(x, y) = x h(y) + y h(x) - h(xy) $$ is continuous on $[0, 1]^2$. We want to show that $F(x, y) > 0$ for $(x, y) \in (0, 1)^2$.

For fixed $y \in (0, 1)$ is $F(0, y) = F(1, y) = 0$, and an elementary calculation shows that $$ \frac{d^2}{dx^2} F(x, y) = - \frac{y(1-y)}{\ln 2 \cdot (1-x)(1-xy)} < 0 $$ for $0 < x < 1$. It follows that $$ x \mapsto F(x, y) $$ is strictly concave on $[0, 1]$ and equal to zero at the endpoints of the interval. This proves that $F(x, y) > 0$ for $0 < x < 1$.