10. (20 points) Let $r_{1}, r_{2}, r_{3}$ be the roots of the real-coefficient equation $x^{3}-x^{2}+a x-b=0$, where $0<r_{i}<1$ $(i=1,2,3)$. Find the maximum possible value of $7 a-9 b$.
10. By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
r_{1}+r_{2}+r_{3}=1 \\
\Rightarrow 1+r_{3}=2-r_{1}-r_{2}=\left(1-r_{1}\right)+\left(1-r_{2}\right) \\
\quad \geqslant 2 \sqrt{\left(1-r_{1}\right)\left(1-r_{2}\right)} .
\end{array}
$$

Similarly, $1+r_{2} \geqslant 2 \sqrt{\left(1-r_{1}\right)\left(1-r_{3}\right)}$,
$$
1+r_{1} \geqslant 2 \sqrt{\left(1-r_{2}\right)\left(1-r_{3}\right)} \text {. }
$$

Multiplying the three inequalities, we get
$$
\begin{array}{l}
\left(1+r_{1}\right)\left(1+r_{2}\right)\left(1+r_{3}\right) \\
\geqslant 8\left(1-r_{1}\right)\left(1-r_{2}\right)\left(1-r_{3}\right) \\
\Rightarrow 1+1+b+a \geqslant 8(1-1+a-b) \\
\Rightarrow 7 a-9 b \leqslant 2,
\end{array}
$$

with equality holding if and only if $r_{1}=r_{2}=r_{3}=\frac{1}{3}$.
Therefore, $a=\frac{1}{3}, b=\frac{1}{27}$.