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.

$\left|y-\frac{4}{3}\right|$ has a maximum at $x=1$ which is $k=\frac{2}{3}$
Derive $f(x)=\frac{x^2+1}{x^2+x+1}-\frac{4}{3}$ and set $f'(x)=0$. Verify that it is a maximum (for instance checking second derivative) then substitute in $\left|y-\frac{4}{3}\right|$ to get $k$.
Hope this helps.
