Show that $\frac c {1+c} \le \frac a {1+a} + \frac b {1+b}$ , for $c \le a+b$ and $a,b,c \ge 0$

Question

So need to show $\frac c {1+c} \le \frac {a+b+2ab} {1+a+b+ab}$

We have $\frac c {1+c} \le \frac {a+b} {1+c}\le \frac {a+b+2ab} {1+c}$

So the whole thing reduces to showing $c \ge a+b+ab$ ?

Can anyone help me with this, I'm rubbish at inequalities...

You should be a bit worried. First, you state that $c\leq a+b$, and then you want to prove that $c\geq a+b+ab$. Since $a$, $b$ and $c$ are non-negative that cannot be true. I suggest that you try to find where things go wrong. — mickep, Jan 19 '15 at 12:22
Possible duplicate of Can I prove this inequality algebraically? — Arnaud D., Dec 04 '19 at 12:46

score 1 · Accepted Answer · answered Jan 19 '15 at 12:32

1

Hint $$\frac c {1+c} \le \frac {a+b+2ab} {1+a+b+ab}\longleftrightarrow c+ac+bc+abc\leq a+b+2ab+ac+bc+2abc$$

answered Jan 19 '15 at 12:32

Farhad

1 Answers1