How to prove an inequality involving logarithm

Question

It is a high school problem my friend asked me.I want to show that for all $0\leq x,y<1$, we have the following inequality: $$(1-y^2)x^2\log(1-y^2)+(1-x^2)y^2\log(1-x^2)\geq 2xy(1-xy)\log(1-xy).$$ It looks symmetric and beautiful on its own and quite strict. I plotted by Mathematica and realized so far that the equality happens when $x=y$ or $x.y=0$. I am not used to inequalities involving logarithms, so if someone can give me some techniques or any methods would be helpful. Thanks!

Answer 1

Divide the inequality by $x^2y^2$ to see that it can be written as $$ f(xy) \le \frac 12 \left( f(x^2) + f(y^2) \right) $$ where $f$ is defined on $ (0, 1)$ as $f(t) = \frac{(1-t)\log(1-t)}{t}$.

With the substitution $x^2 = e^u$, $y^2 = e^v$ this is equivalent to $$ f(e^{(u+v)/2}) \le \frac 12 \left(f(e^u) + f(e^v) \right) $$ for $u, v < 0$, so that it suffices to show that $$ g(u) = f(e^u) = (e^{-u} -1) \log(1-e^u) $$ is a convex function on $(-\infty, 0)$. This can be verified by computing the second derivative.

Spoiler:

$$ g''(u) = e^{-u} \log(1-e^u) + \frac{1}{1-e^u} \ge 0 $$ because of the “well-known” inequality $$ \log(1+x) \ge \frac{x}{1+x} \quad \text{for } x > -1 \, . $$