inequality $\frac{ab}{\ln(e+ab)} \le a + e^b - 1$ for $a,b \ge 0$ For a proof, I tried defining $f(a,b) = \ln(e+ab)(a+e^b-1) - ab$. The values $f(0,b)$ and $f(a,0)$ are both positive, which is good, but the partial derivatives $f_a(a,b)$ and $f_b(a,b)$ are complicated so I seek alternatives.
 A: For case $b=0$ we get that $0 \leq a$ which is true.
For case $a=0$ we get that $0 \leq e^{b}-1$ which means that $b \geq 0$ because $e^b = 1+b +\frac{b^2}{2} + \cdots$ ,also true.
Now for $\frac{a b}{\ln (e +a b)} \leq a +e^b-1$ is stronger than the inequality $\frac{a b}{\ln (e +a b)} \leq a$ because $e^b-1 \geq 0$ is true when $b \geq 0$
Now divide by $a$ since $a >0$ we get that $\frac{b}{\ln( e+ a b) }\leq 1$ multiply by $\ln (e+a b)$ we get that $b \leq \ln(e +a b)$ exponent-ate both sides we get that $e^ b \leq e +a b$ solving for $a$ we get that $a \geq \frac{e^b-e}{b}$ so the inequality is true whenever $a \geq  \frac{e^b-e}{b}$, now we want to prove it for $ a \leq \frac{e^b-e}{b}$.
So we are left with $\frac{ab}{\ln(e+ab)} \leq a+e^b-1$ when $b>0$ and $0<a \leq \frac{e^b-e}{b}$
Since $\ln(e+ab) \geq 1$ exponent-ate both sides we get $e+a b\geq e$ which means that $a b\geq 0$ and this is true.
$\frac{ab}{\ln(e+ab)} \leq ab \leq a+e^b-1$ and since $a>0$ we get that
$\frac{ab}{\ln(e+ab)} \leq ab\leq e^b-1 \leq a+e^b-1$ from the inequalities in the middle we get that $a b \leq e^b-1$ which means that $a \leq \frac{e^b-1}{b}$ which is true since $a \leq \frac{e^b-e}{b}$.
thus completing the proof.
