If $a\geq0$, $b\geq 0$ then the following inequality holds:
$\frac{a}{a+1}+\frac{b}{(a+b+1)^2}+\frac{1}{a+b+1}\le1 $
There are at least three things to try here:
a). Use AM-GM for the denominator of the second fraction, $(a+b)\le \frac{(a+b+1)^2}{4}$
b). Use the fact that $\frac{b}{(a+b+1)^2}\le\frac{b}{(a+b+1)(a+b)}$
c). Consider that $a\geq b$ or $b\geq a$ simply or combined with a). and\or b).
No one of these attempts has brought me to a nice, simple solution and I'm trying to see if there is something around that missed me. This is an inequality given at a high school competition math.
I'd appreciate to receive your valuable feedback in terms of this question. Thanks.