2. [3] Given that $a+b+c=5$ and that $1 \leq a, b, c \leq 2$, what is the minimum possible value of $\frac{1}{a+b}+\frac{1}{b+c}$ ?
Answer: $\frac{4}{7}$ If $a>1$ and $b1$ and $b<2$. Therefore, the sum is minimized when $b=2$. We can then cross-multiply the two fractions and see that we are trying to minimize $\frac{a+c+4}{(a+2)(c+2)}=\frac{7}{(a+2)(c+2)}$. The product of two numbers with a fixed sum is maximized when those two numbers are equal, so $\frac{7}{(a+2)(c+2)}$ is minimized for $a=c=\frac{3}{2}$, which gives us an answer of $\frac{4}{7}$.