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# help

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

Mar 5, 2019

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