Let $a,b,c >0$ Prove the inequality $\displaystyle{\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{1}{a+b+c+1} \geq 1}$

Question

Let $a,b,c >0$

Prove the inequality $\displaystyle{\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{1}{a+b+c+1} \geq 1}$

I dont even know where to begin.

Only interested in hints (not solution)

score 11 · Accepted Answer · answered Mar 15 '12 at 13:05

11

Hint :

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

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

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

answered Mar 15 '12 at 13:05

Pedja

12,883

Duh! I get it now – Kirthi Raman Mar 15 '12 at 13:06
1

The whole expression is $> \frac{a+b+c+1}{a+b+c+1} = 1$ – Kirthi Raman Mar 15 '12 at 13:08

score 3 · Answer 2 · answered Mar 15 '12 at 13:02

3

We have $a+1\leq a+b+c+1$, $b+1\leq a+b+c+1$ and $c+1\leq a+b+c+1$ so $\frac 1{x+1}\geq \frac 1{1+a+b+c}$ where $x=a,b$ or $c$. Now multiply by $x$ to get $\frac x{x+1}\geq \frac x{1+a+b+c}$ and sum.

answered Mar 15 '12 at 13:02

Davide Giraudo

172,925

If $a+1 \geq a+b+c+1$ doesn't that mean $b+c \leq 0$? – Kirthi Raman Mar 15 '12 at 13:05
Sorry I reversed the inequalities, now I think it's correct. – Davide Giraudo Mar 15 '12 at 13:07

Let $a,b,c >0$ Prove the inequality $\displaystyle{\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}+\frac{1}{a+b+c+1} \geq 1}$

2 Answers2