Inequality $\max\left(\frac{1-a^2}{3-2a},\frac{1-b^2}{3-2b}\right)\geq\frac{12}{19}\cdot\frac{3-a^2-2b^2}{6-a-3b}$ Let $0\leq a,b\leq 1$. Prove that
$$\max\left(\frac{1-a^2}{3-2a},\frac{1-b^2}{3-2b}\right)\geq\frac{12}{19}\cdot\frac{3-a^2-2b^2}{6-a-3b}.$$
Equality holds when $a=0$ and $b=2/3$. This seems to be the only case of equality.
The left-hand side is symmetric in $a$ and $b$ but the right-hand side isn't, so we can't simply assume $a\geq b$ or vice-versa. 
A natural thing to try would be to divide into two cases according to which term in the max is greater, but the inequality $$\frac{1-a^2}{3-2a}\geq \frac{1-b^2}{3-2b}$$ doesn't translate easily to nice form of $a$ in terms of $b$.

hint: $$\frac{1-a^2}{3-2a}-\frac{1-b^2}{3-2b}=\frac{(a-b)(2-3a-3b+2ab)}{(2a-3)(2b-3)}$$
