<s>[INST] Proof of three variable equation. The Question: 
Let a,b,c be complex numbers satisfying
$abc = 1$ and $a+b+c =$ $\frac1a + \frac1b + \frac 1c$
Show that at least one of $a,b,c$ must equal $1$.
What I have tried: Rearranging the $RHS$ and subbing in the first equation we get $a+b+c = bc + ac+ab$
Now from Equation 1 we have $ a = \frac{1}{bc}$ and subbing this into the manipulation above we get $\frac{1}{bc} + b+c = bc + \frac 1b + \frac 1c$ 
Now multiplying out by $bc$ we get $1+b+c = (bc)^2 +c + b$ implying that $(bc)^2 = 1$
And from equation one we get $a^2b^2c^2 = 1^2 = 1$ and $b^2c^2 = 1$ therefore $a^2 = 1$ and $a = 1$.
Is this correct/sufficient if not can you point me in the right direction and feel free to show other methods etc. Thanks.
 [/INST] Alternatively, you can just write:
$$(a-1)(b-1)(c-1)=abc - (ab+ac+bc) + (a+b+c) -1 = 1 - [ab+ac+bc-(a+b+c)] - 1 = 0$$
So one of $a-1,b-1,c-1$ must be zero.
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