Equations And Inequalities Question 29
Question: If a, b and c are three positive real numbers such that $ a+b\ge c, $ then
Options:
A) $ \frac{a}{1+a}+\frac{b}{1+b}\ge \frac{c}{1+c} $
B) $ \frac{a}{1+a}+\frac{b}{1+b}<\frac{c}{1+c} $
C) $ \frac{a}{1+a}+\frac{b}{1+b}>\frac{c}{1+c} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] we have
$ \frac{a}{1+a}+\frac{b}{1+b}\ge \frac{a}{1+a+b}+\frac{b}{1+a+b} $
$ =\frac{a+b}{1+a+b}=\frac{1}{\frac{1}{a+b}+1} $
Now, since $ a+b\ge c, $ we get $ \frac{1}{a+b},\le \frac{1}{c}\Rightarrow 1+\frac{1}{a+b}\le 1+\frac{1}{c} $
$ \Rightarrow \frac{1}{\frac{1}{a+b}+1}\ge \frac{1}{\frac{1}{c}+1} $
Thus, $ \frac{a}{1+a}+\frac{b}{1+b}\ge \frac{1}{\frac{1}{c}+1}\ge \frac{1}{\frac{1}{c}+1}=\frac{c}{1+c} $
BETA
