<s>[INST] Solving $\frac{\ 3x}{x^2+x+1}+\frac{2x}{x^2-x+1}=3$ with other approaches 
$$\frac{\ 3x}{x^2+x+1}+\frac{2x}{x^2-x+1}=3$$
$$x=?$$

I solved this problem as follow:
$x=0$ is not a root, we  can divide numerator and denominator of each fraction by $x$:
$$\frac{3}{t+1}+\frac{2}{t-1}=3\quad\quad\text{where $t=x+\frac1x$}$$
$$5t-1=3t^2-3\Rightarrow t=2 , \frac{-1}6$$
Only $x+\frac1x=2$ is acceptable and $x=1$.
I'm looking for other elegant methods to solve this equation.
 [/INST] We have $(x-1)^2\ge 0$ with equality if and only of $x=1$.
But
\begin{align}
(x-1)^2 \ge 0 &\iff x^2-2x+1\ge 0\\
&\iff x^2+x+1\ge 3x\\
&\iff \frac{3x}{x^2+x+1}\le 1
\end{align}
and
\begin{align}
(x-1)^2 \ge 0 &\iff x^2-2x+1\ge 0\\
&\iff x^2-x+1\ge x\\
&\iff \frac{x}{x^2-x+1}\le 1\\
&\iff \frac{2x}{x^2-x+1}\le 2
\end{align}
Then $$\frac{3x}{x^2+x+1} + \frac{2x}{x^2-x+1} \le 3$$
with equality if and only if $x=1$.
</s>