My 100 followers problem!

Relevant wiki: Algebraic Identities

It is given that:

$\begin{aligned} \frac 1a + \frac 1b + \frac 1c = 3 \implies \frac {ab+bc+ca}{abc} = 3 \implies \color{#3D99F6}{ab+bc+ca = 3abc} \end{aligned}$

Now, we have:

$(a+b)ab + (b+c)bc + (c+a)ca \\ = (a+b+\color{#3D99F6}{c})ab + (\color{#3D99F6}{a}+b+c)bc + (c+\color{#3D99F6}{b}+a)ca - \color{#3D99F6}{3abc} \\ = (\color{#3D99F6}{a+b+c})(ab+bc+ca)-3abc \quad \quad \small \color{#3D99F6}{\text{Note that }a+b+c = 1} \\ = \color{#3D99F6}{ab+bc+ca-3abc} \quad \quad \small \color{#3D99F6}{\text{Note that }ab+bc+ca = 3abc} \\ = \boxed{0}$

(1/a)+(1/b)+(1/c)=3

(ab+bc+ca)/abc=3

ab+bc+ca=3abc ---------Eq. 1

We have to find (a+b)(ab)+(b+c)(bc)+(c+a)(ac)

Multiplying and dividing by abc, we get

abc((a+b)/c) + (b+c)/a + (c+a)/b)

From a+b+c=1...

a+b=1-c

b+c=1-a

c+a=1-b

So abc((a+b)/c) + (b+c)/a + (c+a)/b)=

abc((1-c)/c) + (1-a)/a + (1-b)/b)=

abc( 1/a) + (1/b) + (1/c) - 3) =

((abc(ab+bc+ca))/abc) -3abc=

ab+bc+ca-3abc

From Eq. 1 ,3abc = ab+bc+ca

So our answer is (ab+bc+ca) - (ab+bc+ca) = 0

(Sorry for not using latex, I'm still learning latex, plus I'm posting this using my phone).

Yey, new solution

$a+b+c = 1 \implies \color{#D61F06}{a+b} = \color{#3D99F6}{1-c}$ , $\color{#D61F06}{b+c} = \color{#3D99F6}{1-a}$ , $\color{#D61F06}{a+c} = \color{#3D99F6}{1-b}$

$(\color{#D61F06}{a+b})(ab) + (\color{#D61F06}{b+c} )bc + (\color{#D61F06}{a+c})ac = (\color{#3D99F6}{1-c})ab + (\color{#3D99F6}{1-a} )bc + (\color{#3D99F6}{1-b})ac = ab - abc + bc - abc + ca - abc = \color{#BA33D6} {ab+bc+ca - 3abc}$

Now, since $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}= 3$ , multiplying both sides by $abc$ gives us

$abc(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})= 3abc$

$\frac{abc}{a} + \frac{abc}{b} + \frac{abc}{c} = 3abc$

$ab + bc + ca = 3abc \implies \color{#20A900}{ab+bc+ac-3abc} = 0$

Therefore, our answer is $\boxed{0}$ .

$\color{#FFFFFF}{If you read this, congratulations! Either you turned off LaTeX or happened to put your cursor over it}$ .