<s>[INST] Prove that: $\frac{a}{a+1}+\frac{b}{(a+b+1)^2}+\frac{1}{a+b+1}\le1 $ 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.
 [/INST] To prove:$$\frac{a}{a+1}+\frac{b}{(a+b+1)^2}+\frac{1}{a+b+1}\le1$$
Prove:$$\frac{b}{(a+b+1)^2}+\frac{a+b+1}{(a+b+1)^2}\le\frac{1}{a+1}$$
$$\frac{a+2b+1}{(a+b+1)^2}\le\frac{1}{a+1}$$
$$\color{green}{a^2}+\color{blue}{2ab}+\color{red}{a+a+2b+1}\le\color{green}{a^2}+b^2+\color{blue}{2ab}+\color{red}{2a+2b+1}$$
$$0\le b^2$$
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