finding the minimum value of $\frac{x^4+x^2+1}{x^2+x+1}$ given $f(x)=\frac{x^4+x^2+1}{x^2+x+1}$.
Need to find the min value of $f(x)$.
I know it can be easily done by polynomial division but my question is if there's another way
(more elegant maybe) to find the min? 
About my way: $f(x)=\frac{x^4+x^2+1}{x^2+x+1}=x^2-x+1$. (long division)
$x_{min}=\frac{-b}{2a}=\frac{1}{2}$. (when $ax^2+bx+c=0$)
So $f(0.5)=0.5^2-0.5+1=\frac{3}{4}$
Thanks.   

Multiply the fraction by $\frac{x^2-1}{x^2-1}$:$$\frac{(x^2-1)(x^4+x^2+1)}{(x+1)(x-1)(x^2+x+1)}=\frac{x^6-1}{(x+1)(x^3-1)}=\frac{x^3+1}{x+1}=x^2-x+1$$
