Example 13 (2003 Hunan High School Mathematics Competition Question)
Let $x, y$, $z>0$ and $x+y+z=1$, find the minimum value of $f(x, y, z)=\frac{3 x^{2}-x}{1+x^{2}}+$ $\frac{3 y^{2}-y}{1+y^{2}}+\frac{3 z^{2}-z}{1+z^{2}}$.
Given $x, y, z > 0$ and $x + y + z = 1$, we know $0 < x, y, z \leq 1$. First, we prove:
$$\frac{3 x^{2} - x}{1 + x^{2}} \geq \frac{9}{10} x - \frac{3}{10}$$

Since $(*) \Leftrightarrow 10(3 x^{2} - x) \geq (9 x - 3)(x^{2} + 1) \Leftrightarrow 9 x^{3} - 33 x^{2} + 19 x - 3 \leq 0 \Leftrightarrow (3 x - 1)^{2}(x - 3) \leq 0$, which is clearly true, so the inequality $(*)$ holds, with equality if and only if $x = 1 / 3$.

Similarly, we have $\frac{3 y^{2} - y}{1 + y^{2}} \geq \frac{9}{10} y - \frac{3}{10}$, $\frac{3 z^{2} - z}{1 + z^{2}} \geq \frac{9}{10} z - \frac{3}{10}$. Adding these three inequalities, we get $f(x, y, z) = \frac{3 x^{2} - x}{1 + x^{2}} + \frac{3 y^{2} - y}{1 + y^{2}} + \frac{3 z^{2} - z}{1 + z^{2}} \geq \frac{9}{10} - \frac{9}{10} = 0$, with equality if and only if $x = y = z = 1 / 3$. Therefore, the minimum value of $f(x, y, z)$ is 0 when $x = y = z = 1 / 3$.