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.

Let $s=a+b+c = bc + ac+ab$.
Then $a,b,c$ are the roots of
$$(X-a)(X-b)(X-c)=X^3-(a+b+c)X^2+(ab+ac+bc)X-abc$$
$$=X^3-sX^2+sX-1$$
Now all you have to do is observe that $X=1$ is a root of $X^3-sX^2+sX-1$.
