Given that \(a + b + c = 5\) and \(1 \le a,b,c \le 2,\)find the minimum value of \(\frac{1}{a + b} + \frac{1}{b + c}.\)

Guest Mar 5, 2019

#1**+2 **

\(a+b+c=5,~b = 5-a-c\\ \dfrac{1}{a+b} + \dfrac{1}{b+c} = \dfrac{1}{5-c}+\dfrac{1}{5-a}\\ \text{If we want the minimum of this expression it seems clearly that }a,~c\\ \text{should be made as small as possible}\\ \text{Further, as the problem is symmetric in }a \text{ and }c\\ \text{we expect these two to be equal}\\ (a,b,c)=\left(\dfrac 3 2,~2,~\dfrac 3 2\right)\\ \dfrac{1}{a+b}+\dfrac{1}{a+c} =\dfrac 4 7\)

.Rom Mar 6, 2019