TITLE: Prove that: $\frac{a}{a+1}+\frac{b}{(a+b+1)^2}+\frac{1}{a+b+1}\le1 $
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: Let $$f(a,b)=\frac{a}{a+1}+\frac{b}{(a+b+1)^2}+\frac{1}{a+b+1}.$$
Observe that, for all  $a\geq0$, $f(a,0)=1$. It is thus enough to show that, for any fixed $a\geq0$, the function $b\mapsto f(a,b)$ is decreasing. We thus compute its derivative $$0+\frac{1}{(a+b+1)^2}-\frac{2b}{(a+b+1)^3}-\frac{1}{(a+b+1)^2}.$$ This derivative is negative, thus the function $b\mapsto f(a,b)$ is decreasing on $[0,+\infty)$, thus we have the inequality $$\frac{a}{a+1}+\frac{b}{(a+b+1)^2}+\frac{1}{a+b+1}\leq f(a,0)=1$$ for all $a\geq0$ and $b\geq0$.