Show that $\frac{a}{a+1}+\frac{b}{b+1}\geq\frac{a+b}{1+a+b}\geq\frac{c}{1+c}$ whenever $c\leq a+b$

Question

I found a post on ME here. It sais:

Try to show that $$\frac{a}{a+1}+\frac{b}{b+1}\geq\frac{a+b}{1+a+b}\geq\frac{c}{1+c}$$ whenever $c\leq a+b$

I wasn't able to solve this. How does it work?

I can give an answer with calculus. Is that preferable, or should I stick to ordinary AM-GM etc. — Sarvesh Ravichandran Iyer, Nov 18 '17 at 03:33
See: Given positive real numbers $a, b, c$ with $a<b+c$, show that $a/(1+a)<b/(1+b)+c/(1+c)$ — Ben, Nov 18 '17 at 04:41
Just study the function $x\to \dfrac{x}{1+x}.$ (Draw the graph. Look at the derivative, concavity and....) — Bumblebee, Nov 18 '17 at 05:03

score 2 · Accepted Answer · edited Nov 18 '17 at 04:56

We know that,

$$\frac{a}{1+a} \ge \frac{a}{1+a+b} $$

and

$$\frac{b}{1+b} \ge \frac{b}{1+a+b} $$

Adding the above two equations. $$\frac{a}{1+a} + \frac{b}{1+b} \ge \frac{a}{1+a+b} + \frac{b}{1+a+b} $$

$$\frac{a}{1+a} + \frac{b}{1+b}\ge \frac{a+b}{1+a+b} $$

Now assume $a+b = k$

Given

\begin{align} & a+b \ge c \\ &\Rightarrow k \ge c \\ &\Rightarrow k + 1 \ge c + 1 \\ &\Rightarrow \frac{1}{k + 1} \le \frac{1}{c+ 1} \\ &\Rightarrow -\frac{1}{k + 1} \ge -\frac{1}{c + 1} \\ &\Rightarrow 1-\frac{1}{k + 1} \ge 1 -\frac{1}{c + 1} \\ &\Rightarrow \frac{k}{k + 1} \ge \frac{c}{c + 1} \\&\Rightarrow \frac{a+b}{1 + a+ b} \ge \frac{c}{c + 1} \end{align}

Hence Proved

$$\frac{a}{1+a} + \frac{b}{1+b}\ge \frac{a+b}{1+a+b} \ge \frac{c}{c + 1}$$

This is dope. Thank you very much! +500 ;-) – mangolassi Nov 18 '17 at 12:29 — mangolassi, Nov 18 '17 at 12:29

Show that $\frac{a}{a+1}+\frac{b}{b+1}\geq\frac{a+b}{1+a+b}\geq\frac{c}{1+c}$ whenever $c\leq a+b$

1 Answers1