Minimum value of given expression

let y=$\frac{x^2+x+1}{x^2-x+1}$

=>$x^2(y-1)-x(y+1)+(y-1)=0$

As x is real, the discriminant= $(y+1)^2-4(y-1)^2≥0$

=>$(y-3)(y-\frac{1}{3})≤0$

=>$\frac{1}{3}≤y≤3$

Here is an 'algebra solution':

$\frac {x^2 + x + 1 } {x^2 - x + 1 } = \frac{(x+1)^2-(x+1)+1}{(x+1)^2-3(x+1)+3} = \frac{1}{3} + \frac{2}{3} \frac{(x+1)^2}{x^2-x+1} = \frac{1}{3} + \frac{2}{3} \frac{(x+1)^2}{(x-\frac{1}{2})^2+\frac{3}{4}}$.

Since the last term is greater than zero when $x\neq -1$, we see that the minimum is

$\frac{1}{3}$.

Of course, this is cheating since I know the answer from J.D.'s solution and this suggests the way to expand the expression.

Hint: Take the derivative w.r.t $x$ and equate it with zero, you get: $$ \frac{d}{dx} \frac {x^2 + x + 1 } {x^2 - x + 1 } = - \frac {2(x^2 - 1)} {(x^2 - x + 1)^2} = 0.$$ So $x = \pm 1$ at the extrema. Test both for minimum.

Edit: this tutorial page might be helpful.