# 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.

• d). Simply move things around and enjoy :p Aug 21, 2012 at 10:06

To prove:$$\frac{a}{a+1}+\frac{b}{(a+b+1)^2}+\frac{1}{a+b+1}\le1$$ Prove:$$\frac{b}{(a+b+1)^2}+\frac{a+b+1}{(a+b+1)^2}\le\frac{1}{a+1}$$ $$\frac{a+2b+1}{(a+b+1)^2}\le\frac{1}{a+1}$$ $$\color{green}{a^2}+\color{blue}{2ab}+\color{red}{a+a+2b+1}\le\color{green}{a^2}+b^2+\color{blue}{2ab}+\color{red}{2a+2b+1}$$ $$0\le b^2$$

• IMHO best solution. Simple and obvious. Aug 21, 2012 at 12:29

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 $b\geq0$.

• thank you for your solution (+1). Aug 21, 2012 at 11:10
• Got it. Nice solution! Aug 21, 2012 at 11:13

Alternatively you can use b') $\frac{b}{(a+b+1)^2}\le\frac{b}{(a+b+1)(a+1)}$. If you insert it, the LHS is $1$.

• @HenrikRueping: this is the solution I'm looking for. Superb! (+1) One may simply solve it this way just at sight. Aug 21, 2012 at 11:05