Question

If $a+b+c=6,\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2},\mathrm{then}\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}=\_\_\_\_\_\_\_\_\_\_.$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2}...\left(1\right) \\ a+b+c=6...\left(2\right) \\ \mathrm{Now}, \\ \frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b} \\ \mathrm{Adding}\mathrm{and}\mathrm{subtracting}3,\mathrm{we}\mathrm{get} \\ =\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+3-3 \\ =\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1+1+1-3 \\ =\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+\frac{a}{a}+\frac{b}{b}+\frac{c}{c}-3 \\ =\left(\frac{a}{a}+\frac{b}{a}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{b}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{b}{c}+\frac{c}{c}\right)-3 \\ =\left(\frac{a+b+c}{a}\right)+\left(\frac{a+b+c}{b}\right)+\left(\frac{a+b+c}{c}\right)-3 \\ =\left(\frac{6}{a}\right)+\left(\frac{6}{b}\right)+\left(\frac{6}{c}\right)-3\left(\mathrm{From}\left(2\right)\right) \\ =\frac{6}{a}+\frac{6}{b}+\frac{6}{c}-3 \\ =6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-3 \\ =6\left(\frac{3}{2}\right)-3\left(\mathrm{From}\left(1\right)\right) \\ =9-3 \\ =6 \end{gathered}$$

Hence, if $a+b+c=6,\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2},\mathrm{then}\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}=\underline{6}.$