34. Given the equation $x^{3}-a x^{2}+b x-c=0$ has three positive roots (which can be equal), find the minimum value of $\frac{1+a+b+c}{3+2 a+b}-\frac{c}{b}$. (2008 Turkey Training Team Problem)
34. Considering the equation $x^{3}-a x^{2}+b x-c=0$ has three equal positive roots $p, q, r$, when $\frac{1+a+b+c}{3+2 a+b}-\frac{c}{b}=\frac{1}{3}$, it is only necessary to prove $\frac{1+a+b+c}{3+2 a+b}-\frac{c}{b} \geqslant \frac{1}{3}$. This is equivalent to proving $a b+$ $2 b^{2} \geqslant 9 c+6 a c$.

By Vieta's formulas, we have $a=p+q+r, b=p q+q r+r p, c=p q r$. By the AM-GM inequality, we get $(p+q+r)(p q+q r+r p) \geqslant 9 p q r$, i.e., $a b \geqslant 9 c$. Also, by the AM-GM inequality, we get $(p q+q r+$ $r p)^{2} \geqslant 3 \operatorname{pqr}(p+q+r)$. i.e., $b^{2} \geqslant 3 a c$. Therefore, the inequality holds.