Let $P(x)$ be a polynomial such that $P(x)=x^3-2x^2+bx+c$. Roots of $P(x)$ belong to interval $(0,1)$. Prove that $8b+9c \leq 8$. When does equality hold?
1. Given the polynomial \( P(x) = x^3 - 2x^2 + bx + c \), we know that its roots \( x_1, x_2, x_3 \) lie in the interval \( (0, 1) \).
2. By Vieta's formulas, we have:
   \[
   x_1 + x_2 + x_3 = 2,
   \]
   \[
   x_1 x_2 + x_2 x_3 + x_3 x_1 = b,
   \]
   \[
   x_1 x_2 x_3 = -c.
   \]
3. We need to prove that \( 8b + 9c \leq 8 \). Substituting the expressions for \( b \) and \( c \), the inequality becomes:
   \[
   8(x_1 x_2 + x_2 x_3 + x_3 x_1) - 9x_1 x_2 x_3 \leq 8.
   \]
4. To homogenize the inequality, we multiply both sides by \( (x_1 + x_2 + x_3) \):
   \[
   8(x_1 x_2 + x_2 x_3 + x_3 x_1)(x_1 + x_2 + x_3) - 9x_1 x_2 x_3(x_1 + x_2 + x_3) \leq 8(x_1 + x_2 + x_3).
   \]
5. Since \( x_1 + x_2 + x_3 = 2 \), we substitute this into the inequality:
   \[
   8(x_1 x_2 + x_2 x_3 + x_3 x_1) \cdot 2 - 9x_1 x_2 x_3 \cdot 2 \leq 8 \cdot 2.
   \]
   Simplifying, we get:
   \[
   16(x_1 x_2 + x_2 x_3 + x_3 x_1) - 18x_1 x_2 x_3 \leq 16.
   \]
6. Dividing the entire inequality by 2:
   \[
   8(x_1 x_2 + x_2 x_3 + x_3 x_1) - 9x_1 x_2 x_3 \leq 8.
   \]
7. This is the same as our original inequality. To prove it, we use Schur's inequality, which states:
   \[
   x_1^3 + x_2^3 + x_3^3 + 3x_1 x_2 x_3 \geq x_1^2 x_2 + x_1 x_2^2 + x_2^2 x_3 + x_2 x_3^2 + x_3^2 x_1 + x_3 x_1^2.
   \]
8. We need to show that:
   \[
   x_1^3 + x_2^3 + x_3^3 + 3x_1 x_2 x_3 \geq x_1^2 x_2 + x_1 x_2^2 + x_2^2 x_3 + x_2 x_3^2 + x_3^2 x_1 + x_3 x_1^2.
   \]
   This inequality holds by Schur's inequality.

9. For equality to hold in Schur's inequality, the roots must be equal. Given \( x_1 + x_2 + x_3 = 2 \) and \( x_1, x_2, x_3 \in (0, 1) \), equality is achieved if and only if \( x_1 = x_2 = x_3 = \frac{2}{3} \).

The final answer is \( \boxed{ 8b + 9c \leq 8 } \). Equality holds when \( x_1 = x_2 = x_3 = \frac{2}{3} \).