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 

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. 
