### Instruction: Given that $a + b + c = 5$ and $1 \le a,$ $b,$ $c \le 2,$ find the minimum value of
\[\frac{1}{a + b} + \frac{1}{b + c}.\].### Response: I want to minimize the expression, so I think about what makes a fraction smaller.
The numerator is fixed at 1, so the fraction gets smaller as the denominator gets larger.
So I want to make $a + b$ and $b + c$ as large as possible, subject to the constraints.
Since $a,$ $b,$ $c$ are all between 1 and 2, the largest they can be is 2.
So I try to make $a,$ $b,$ $c$ all equal to 2.
But that doesn't work, because then $a + b + c = 6,$ which violates the given equation.
So I have to lower one of the variables to 1, and keep the other two at 2.
Which one should I lower?
Well, if I lower $a$ to 1, then $a + b = 3$ and $b + c = 4.$
If I lower $b$ to 1, then $a + b = 3$ and $b + c = 3.$
If I lower $c$ to 1, then $a + b = 4$ and $b + c = 3.$
The last option seems to give me the largest denominators, so I choose $a = 2,$ $b = 2,$ $c = 1.$
Then the expression becomes
\[\frac{1}{2 + 2} + \frac{1}{2 + 1} = \frac{1}{4} + \frac{1}{3}.\]
I can simplify this by finding a common denominator of 12.
Then I get
\[\frac{3}{12} + \frac{4}{12} = \frac{7}{12}.\]