If $x$ is real, evaluate $k$ in absolute inequality 
If $x$ is real and $$y=\frac{(x^2+1)}{x^2+x+1}$$ then it can be shown that $$\left|y-\frac{4}{3}\right|\leq k$$ Evaluate $k$

My attempt,
\begin{align}\left(y-\frac{4}{3}\right)^2&\leq k^2\\
\sqrt{\left(y-\frac{4}{3}\right)^2}&\leq k\end{align}
I don't know how to proceed anymore. Thanks in advance.

$$\frac{x^2+1}{x^2+x+1}-\frac 43=1-\frac x{x^2+x+1}-\frac 43=-\frac 13-\frac x{x^2+x+1}$$
So you need to find $m$ such that $$\frac{x}{x^2+x+1}\leq m$$
Since both $m,x^2+x+1>0$ we can multiply it out we get $$ mx^2+(m-1)x+m\geq 0$$
Now we must have that $m>0$ and that $\Delta\leq 0$(discriminant), now just find the minimal $m$ that satisfies those requirements and then the minimum will be $$k=\frac 13+m$$
I'll leave it to you to find the $k$.
