<s>[INST] 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.
 [/INST] If $x=-1$ then $\left|\frac{x^2+1}{x^2+x+1}-\frac{4}{3}\right|=\frac{2}{3}$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$\left|\frac{x^2+1}{x^2+x+1}-\frac{4}{3}\right|\leq\frac{2}{3}$$ or
$$-\frac{2}{3}\leq\frac{x^2+1}{x^2+x+1}-\frac{4}{3}\leq\frac{2}{3}$$ or
$$\frac{2}{3}\leq\frac{x^2+1}{x^2+x+1}\leq2,$$
which is true because the left inequality it's $(x-1)^2\geq0$
and the right inequality it's $(x+1)^2\geq0$.
Done!
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