Prove that $\frac{c}{c+1}< \frac{a}{a+1} + \frac{b}{b+1}$

Question

Let $a$, $b$ and $c$ be positive real numbers such that $c < a+b$. Prove that $$\frac{c}{c+1}< \frac{a}{a+1} + \frac{b}{b+1}.$$

Answer 1

Multiplying by $(a+1)(b+1)(c+1)$ we get

$$ c(a+1)(b+1) < (c+1)(2ab + a + b) $$

$$ cab + cb + ca + c < 2cab + ca + cb + 2ab + a + b$$

$$ c < cab + 2ab + a + b $$

And since we know $c<a+b$, the above is true, because $a,b,c > 0$ so that $cab + 2ab > 0$

Answer 2

You can use this $\ \dfrac xy<\dfrac uv\implies \dfrac{x}{y}<\dfrac{x+u}{y+v}\ $ for positive denominators.

(see for instance Show that a simple inequality holds.)

In particular $\dfrac{c}{a+b}<\dfrac 11\implies\dfrac{c}{a+b}<\dfrac{c+1}{a+b+1}\implies\dfrac{c}{c+1}<\dfrac{a+b}{a+b+1}$

Now use that $\begin{cases}a+b+1\ge a+1\\a+b+1> b+1\end{cases}$ since $a,b> 0$

So $\dfrac{c}{c+1}<\dfrac{a}{a+b+1}+\dfrac{b}{a+b+1}<\dfrac{a}{a+1}+\dfrac{b}{b+1}$

Answer 3

If $c \leq a$, then we have $\frac c{c + 1} \leq \frac a{a + 1} < \frac a {a + 1} + \frac b {b + 1}$. Same if $c \leq b$.

Suppose $c > a$ and $c > b$. Then we have $\frac c{c + 1} < \frac{a + b}{c + 1} = \frac a{c + 1} + \frac b {c + 1} < \frac a{a + 1} + \frac b{b + 1}$.