Possible values of $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ given $x+y+z=1$ Given $x,y,z$ are real numbers and $x+y+z=1$ and $x$ is not equal to $z$, if $ {1\over x} + {1\over y} + {1\over z} = m $, which of the following values of $m$ are possible?
(A) 1
(B) 2 
(C) 3 
(D) All of these.
I don't know where to start. I thought of taking $x=y$ but there doesn't seem to be any pattern to obtain the answer.

(1) Consider the function $f(x)=x^3-x^2-2x+2=(x-1)(x^2-2)$ we have three real roots and $x+y+z=-{1\over -1}=1,{1\over x}+{1\over y}+{1\over z}=-{2\over-2}=1$.
(2) Consider the function $f(x)=x^3-x^2-40x+20$. Set the derivative equal $0$ we get $x=4$ and $x=-{10\over3}$ are two local extrema.  When $x=4, f(x)<0$, when $x=-{10\over 3}, f(x)>0$ so there are three real roots. We have $x+y+z=1$ and ${1\over x}+{1\over y}+{1\over z}=-{-40\over20}=2$.
(3) Consider the function $f(x)=x^3-x^2-40x+{40\over3}$. We still have the same local extrema as (2) and when $x=4, f(x)<0$ and when $x=-{10\over 3}, f(x)>0$ so there are three real roots. We have $x+y+z=1$ and ${1\over x}+{1\over y}+{1\over z}=-{-40\over{40\over3}}=3$.
