Minimum value of given expression What is the minimum value of the $$ \frac {x^2 + x + 1 }  {x^2 - x + 1 } \ ?$$
I have solved by equating it to m and then discriminant greater than or equal to zero and got the answer, but can algebraic manipulation is possible 
 A: 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= $(y+1)^2-4(y-1)^2≥0$ 
=>$(y-3)(y-\frac{1}{3})≤0$
=>$\frac{1}{3}≤y≤3$
A: 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.
A: 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.
A: For $x\geq0$ we have $$\frac{x^2+x+1}{x^2-x+1}=1+\frac{2x}{x^2-x+1}\geq1.$$
For $x<0$ by AM-GM we obtain:
$$\frac{x^2+x+1}{x^2-x+1}=1+\frac{2x}{x^2-x+1}=1+\frac{2}{x+\frac{1}{x}-1}=$$
$$=1-\frac{2}{-x+\frac{1}{-x}+1}\geq1-\frac{2}{2\sqrt{-x\cdot\frac{1}{-x}}+1}=\frac{1}{3}.$$
The equality occurs for $-x=\frac{1}{-x}$ or for $x=-1,$ which says that we got a minimal value.
By the same way we can get a maximal value if you want. 
