12. Given that $a, b, c$ are positive real numbers, and $a+b+c=1$, find the minimum value of $u=\frac{3 a^{2}-a}{1+a^{2}}+\frac{3 b^{2}-b}{1+b^{2}}+\frac{3 c^{2}-c}{1+c^{2}}$.
12. Consider the function $f(x)=\frac{x}{1+x^{2}}, x \in(0,1)$. It is easy to prove that for $x_{1}, x_{2} \in(0,1)$, we have $\left(x_{1}-x_{2}\right) \cdot\left[f\left(x_{1}\right) -f\left(x_{2}\right)\right] \geqslant 0$. Therefore, we have $\left(a-\frac{1}{3}\right)\left[f(a)-f\left(\frac{1}{3}\right)\right] \geqslant 0$, which means $\left(a-\frac{1}{3}\right)\left[\frac{a}{1+a^{2}}-\frac{3}{10}\right] \geqslant 0$.
Thus, $\frac{a^{2}-\frac{1}{3} a}{1+a^{2}} \geqslant \frac{3}{10}\left(a-\frac{1}{3}\right)$, or $\frac{3 a^{2}-a}{1+a^{2}} \geqslant \frac{9}{10}\left(a-\frac{1}{3}\right)$,
Similarly: $\frac{3 b^{2}-b}{1+b^{2}} \geqslant \frac{9}{10}\left(b-\frac{1}{3}\right), \frac{3 c^{2}-c}{1+c^{2}} \geqslant \frac{9}{10}\left(c-\frac{1}{3}\right)$.
Therefore, $u \geqslant \frac{9}{10}\left(a-\frac{1}{3}+b-\frac{1}{3}+c-\frac{1}{3}\right)=0$, with equality when $a=b=c=\frac{1}{3}$. Hence, the minimum value of $u$ is 0.
Note: If we construct the function $f(x)=\frac{3 x^{2}-x}{1+x^{2}}, x \in(0,1)$, and then try to find the maximum value of $f(x)$, it would be very difficult to obtain the result.