9. Let $x, y$ be positive real numbers, and $x+y=1$. Then the minimum value of $\frac{x^{2}}{x+2}+\frac{y^{2}}{y+1}$ is $\qquad$ .
9. $\frac{1}{4}$.

By Cauchy-Schwarz inequality, we have
$$
\frac{x^{2}}{x+2}+\frac{y^{2}}{y+1} \geqslant \frac{(x+y)^{2}}{x+2+y+1}=\frac{1}{4} \text {. }
$$

Equality holds if and only if $\frac{\frac{x^{2}}{x+2}}{x+2}=\frac{\frac{y^{2}}{y+1}}{y+1} \Leftrightarrow x=2 y$, i.e., $y=\frac{1}{3}, x=\frac{2}{3}$.