The Other Way Around

$a^{-1}+b^{-1}+c^{-1}$ can be written as $\frac { ab+ac+bc }{ abc }$

And we have $ab+ac+bc=\frac { 1 }{ 2 } \left[ { \left( a+b+c \right) }^{ 2 }-({ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }) \right]$

$\Rightarrow$ $ab+ac+bc=\frac { 1 }{ 2 } (1-2)$ $\Rightarrow$ $ab+bc+ac=\frac{-1}{2}$

From the identity

${ a }^{ 3 }+{ { b }^{ 3 }+c }^{ 3 }-3abc=(a+b+c)\left[ { (a^{2}+b^{2}+c^{2}) }-(ab+ac+bc) \right]$

$\Rightarrow$ $3-3abc=1(2+\frac{1}{2})$ $\Rightarrow$ $abc=\frac{1}{6}$

Therefore: $a^{-1}+b^{-1}+c^{-1}$ $=$ $\frac { ab+ac+bc }{ abc }$ $=\frac{\frac{-1}{2}}{\frac{1}{6}}$

$\Rightarrow$ $a^{-1}+b^{-1}+c^{-1}=-3$

The first thing that I did was rewrite the problem. $a^{-1} + b^{-1} + c^{-1} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc}{abc} + \frac{ac}{abc} + \frac{ab}{abc} = \frac{bc+ac+ab}{abc}$ Next, I manipulated the original equation to find a value equivalent to bc+ac+ab and another one equivalent to abc. $a+b+c = 1$ $(a+b+c)^{2} = 1^{2}$ $(a+b+c)(a+b+c) = 1$ $a(a+b+c) + b(a+b+c) + c(a+b+c) = 1$ $a^{2} + ab + ac + ab + b^{2} + bc + ac + bc + c^{2} = 1$ $a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac = 1$ $(a^{2} + b^{2} + c^{2}) + 2ab + 2bc + 2ac = 1$ The expression in parenthesis is the second equation we were given; therefore, $2 + 2ab + 2bc + 2ac = 1$ $2ab + 2bc + 2ac = -1$ $ab + bc + ac = -\frac{1}{2}$ There, we have the expression for the numerator. The denominator is a little more involved. $a+b+c = 1$ $(a+b+c)^{3} = 1^{3}$ $(a+b+c)(a+b+c)^{2} = 1$ From the previous part we know that, $(a+b+c)^{2} = 2 + 2ab + 2bc + 2ac$ Therefore, $(a+b+c)(2+2ab+2bc+2ac) = 1$ $a(2+2ab+2bc+2ac) + b(2+2ab+2bc+2ac) + c(2+2ab+2bc+2ac) = 1$ $2a+2a^{2}b+2abc+2a^{2}c+2b+2ab^{2}+2b^{2}c+2abc+2c+2abc+2bc^{2}+2ac^{2} = 1$ Combine like terms and strategically place parentheses: $6abc + (2a+2b+2c) + (2a^{2}c+2a^{2}b) + (2ab^{2}+2b^{2}c) + (2bc^{2}+2ac^{2}) = 1$ $6abc + 2(a+b+c) + 2a^{2}(b+c) + 2b^{2}(a+c) + 2c^{2}(a+b) = 1$ $6abc + 2(a+b+c) + 2a^{2}(a+b+c-a) + 2b^{2}(a+b+c-b) + 2c^{2}(a+b+c-c) = 1$ Substitute 1 for every a+b+c that appears (first equation given). $6abc + 2(1) + 2a^{2}(1-a) + 2b^{2}(1-b) + 2c^{2}(1-c) = 1$ $6abc + 2 + 2a^{2} - 2a^{3} + 2b^{2} - 2b^{3} + 2c^{2} - 2c^{3} = 1$ $6abc + 2 + 2a^{2} + 2b^{2} + 2c^{2} - 2a^{3} - 2b^{3} - 2c^{3} = 1$ $6abc + 2 + 2(a^{2} + b^{2} + c^{c}) - 2(a^{3} + b^{3} + c^{3}) = 1$ Take a look at the second and third given equations and substitute. $6abc + 2 + 2(2) - 2(3) = 1$ $6abc + 2 + 4 - 6 = 1$ $6abc = 1$ $abc = \frac{1}{6}$ There, we have an expression for the denominator. Now we plug them back in. $\frac{bc+ac+ab}{abc} = \frac{-\frac{1}{2}}{\frac{1}{6}} = \boxed{-3}$

Squaring the first equation: $a^2+b^2+c^2+2(ab+ac+bc)$

but $a^2+b^2+c^2=2$ .

So $ab +bc + ac =- \frac {1} {2}$ [ $4^{th}$ equation]

Multiplying the first and second equation: $a^3+b^3+c^3+ab^2+ac^2+a^2b+bc^2+a^2c+b^2c=2$

but $a^3+b^3+c^3=3$

so $ab^2+ac^2+a^2b+bc^2+a^2c+b^2c=-1$ [ $5^{ th}$ equation]

Multiplying the first and fourth equation: $ab^2+ac^2+a^2b+bc^2+a^2c+b^2c+3abc=-\frac {1} {2}$

but $ab^2+ac^2+a^2b+bc^2+a^2c+b^2c=-1$

so $abc=\frac {1} {6}$

$\boxed {a^{-1}+ b^{-1}+c{-1}=}\frac {1} {a} +\frac {1} { b}+\frac {1} {c}=\frac {bc+ ac+ab} {abc}=\frac {-\frac {1} {2}} {\frac {1} {6}}=\boxed {-3}$

We have to only remember two formulas i.e. $[a+b+c]^2$ = $a^2$ + $b^2$ + $c^2$ + 2(ab +bc +ca) and $a^3$ + $b^3$ + $c^3$ - 3abc = (a + b + c)( $a^2$ + $b^2$ + $c^2$ - ab -bc -ca). By these two formulas we can easily equate the answer.

Subtracting the 1st equation from the 2nd:

$a(a-1)+b(b-1)+c(c-1) = 1$ --- ( 1 )

Subtracting the 2nd equation from the 3rd:

$a^{2}(a-1)+b^{2}(b-1)+c^{2}(c-1) = 1$ --- ( 2 )

From equation 1, three equations can be deduced:

1. $a-1 = -(b+c)$
2. $b-1 = -(a+c)$
3. $c-1 = -(a+b)$

Doing the necessary substitutions with the equations above into equation ( 1 ) and ( 2 ), we get:

$a(b+c)+b(a+c)+c(a+b) = -1$ --- ( a ) $a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b) = -1$ --- ( b )

Subtracting the equation (a) from (b), expanding, cleaning up and factorising:

$6abc+a^{2}(b+c)+b^{2}(a+c)+c^{2}(a+b) = 0$

Therefore,

$abc = \frac{1}{6}$

Now with the 1st equation, by doing the necessary multiplications we can get

$ab+b^{2}+bc = b$

$a^{2}+ab+ac = a$

$ac+bc+c^{2} = c$

Adding up and simplifying,

$a+b+c = a^{2}+b^{2}+c^{2} +2(ab+bc+ac)$

Which reduces to $ab+bc+ac = \frac{-1}{2}$

Now we have all the information that we need. The equation that we want to find the value of can be written as

$\frac{ab+ac+bc}{abc}$

Thus the value is $\boxed{-3}$

Because

ab+bc+ca = [(a+b+c)^2-(a^2+b^2+c^2)]/2 = [1^2-2]/2 = -1/2

and

a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)

<=> 3-3abc = 1*(2+1/2),

<=> abc = 1/6,

it follows that

1/a+1/b+1/c

= (ab+bc+ca)/(abc)

= (-1/2)/(1/6)

= -3.

I wanted to use Newton's sums in order to solve this problem. So, let's define: $e_1 \equiv a + b + c\;, \qquad e_2 \equiv ab + bc + ca\;, \qquad e_3 \equiv abc\;.$ The first three Newton's sums, the ones given by the problem, are thus: $p_1 \equiv a + b + c = e_1\;,\\ p_2 \equiv a^2 + b^2 + c^2 = e_1p_1 - 2e_2\;,\\ p_3 \equiv a^3 + b^3 + c^3 = e_1p_2 - e_2p_1 + 3e_3\;.$ Since $p_1 = 1\;, \qquad p_2 = 2\;, \qquad p_3 = 3\;,$ we can easily determine $e_1 = 1\;, \qquad e_2 = -\frac{1}{2}\;, \qquad e_3 = \frac{1}{6}\;.$ Now, $a^{-1} + b^{-1} + c^{-1} = \frac{e_2}{e_3}\;,$ and so we find the result $a^{-1} + b^{-1} + c^{-1} = -3\;,$

${\bf 1 = (a + b + c)^2 = 2 + 2(ab + ac + bc) \implies \frac{-1}{2} = ab + ac + bc }$

${\bf 1 = (a + b + c)^3 = 3 + 3(ab + ac + bc)(a + b + c) - 3abc \implies }$

${\bf -2 = \frac{-3}{2} - 3abc \implies }$

${\bf -4 = -3 - 6abc \implies abc = \frac{1}{6} }$

${\bf \therefore \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc + ac + ab}{abc} = -3. }$