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
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, \[\frac{a}{1+a}+\frac{b}{1+b}\ge \frac{1}{\frac{1}{c}+1}\ge \frac{1}{\frac{1}{c}+1}=\frac{c}{1+c}\] |
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