How Quickly Can You Solve This?

First note that it is safe to assume that none of $a,b,c$ equal $0.$ Now let the given expression be $S.$ Then we have that

$S + 3 = S + \dfrac{a}{a} + \dfrac{b}{b} + \dfrac{c}{c} = \left(\dfrac{b + c}{a} + \dfrac{a}{a}\right) + \left(\dfrac{c + a}{b} + \dfrac{b}{b}\right) + \left(\dfrac{a + b}{c} + \dfrac{c}{c}\right) =$

$\dfrac{a + b + c}{a} + \dfrac{a + b + c}{b} + \dfrac{a + b + c}{c} = (a + b + c)\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) = 6 \times 2 = 12.$

Thus $S + 3 = 12,$ and so $S = \boxed{9}.$

$\large\color{#D61F06}{a}+\color{#69047E}{b}+\color{#3D99F6}{c} = 6$

$\large\frac{1}{\color{#D61F06}{a}} + \frac{1}{\color{#69047E}{b}} + \frac{1}{\color{#3D99F6}{c}} = 2$

Notice that each denominator has the other 2 variables as the nominator, that would be achievable only by multiplication

So multiply the first 2 equations

$\large(\color{#D61F06}{a}+\color{#69047E}{b}+\color{#3D99F6}{c})\times(\frac{1}{\color{#D61F06}{a}} + \frac{1}{\color{#69047E}{b}} + \frac{1}{\color{#3D99F6}{c}}) = (6\times2)$

$\large\color{#D61F06}{\frac{a}{a}}+\color{#69047E}{\frac{b}{b}}+\color{#3D99F6}{\frac{c}{c}}+\frac{\color{#69047E}{b}+\color{#3D99F6}{c}}{\color{#D61F06}{a}} + \frac{\color{#D61F06}{a}+\color{#3D99F6}{c}}{\color{#69047E}{b}} + \frac{\color{#D61F06}{a}+\color{#69047E}{b}}{\color{#3D99F6}{c}} = 12$

$\large\frac{\color{#69047E}{b}+\color{#3D99F6}{c}}{\color{#D61F06}{a}} + \frac{\color{#D61F06}{a}+\color{#3D99F6}{c}}{\color{#69047E}{b}} + \frac{\color{#D61F06}{a}+\color{#69047E}{b}}{\color{#3D99F6}{c}} = 12-3=9$

We are given that a+b+c=6. So, $\frac{1}{a+b+c}=\frac{1}{6}.$ Another information we are given is $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2.$ Relating these two, we can say that $\frac{12}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ , that is, $\frac{12}{a+b+c}=\frac{bc+ac+ab}{abc}$ . $12abc=(a+b+c)(bc+ac+ab)$ . $12abc=abc+a^{2}c+a^{2}b+b^{2}c+abc+ab^{2}+bc{2}+ac^{2}+abc.$ $12abc=a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b)+3abc.$ $a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b)=9abc.$ $\frac{a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b)}{abc}=9.$ $\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=9.$

We can assume a, b, c to be the sides of a triangle since its logical that a, b, c are positive. Now make use of semi-perimeter of a triangle.

Multiplying the two given equations then simplifying. You'll get the answer. PROMISE.

a ┼ b ┼ c = 6

1/a ┼ 1/b ┼ 1/c = 2

(b┼c)/a ┼ (c┼a)/b ┼ (a┼b)/c = ?

b/a ┼ c/a ┼ c/b ┼ a/b ┼ a/c ┼ b/c

├ (a ┼b ┼c)/a - 1 ┼ (a ┼b ┼c)/b - 1 ┼ (a ┼b ┼c)/c- 1

├ 6/a - 1 ┼ 6/b - 1 ┼ 6/c - 1

├ 6 × (1/a ┼ 1/b ┼ 1/c) - 3

6 × (2) -3 = 9

Write (a+b+c-a)/a as first term, =(6/a)-1. Expn becomes 6(E(1/a)-3. E(1/a) is given as 2.E stands for sigma. Hence answer is 9.