23. 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)
23. Let the three positive roots (which can be equal) of the equation $x^{3}-a x^{2}+b x-c=0$ be $p, q, r$. By Vieta's formulas, we have $a=p+q+r, b=p q+q r+r p, c=p q r$, so
$$\begin{array}{c}
\frac{1+a+b+c}{3+2 a+b}=\frac{(1+p)(1+q)(1+r)}{(1+p)(1+q)+(1+q)(1+r)+(1+r)(1+p)} \\
\frac{c}{b}=\frac{p q r}{p q+q r+r p}
\end{array}$$

Let $f(x, y, z)=\frac{x y z}{x y+y z+z x}$. The problem is to find the minimum value of $f(p+1, q+1, r+1)-f(p, q, r)$.

Assume without loss of generality that $p \geqslant q \geqslant r>0$. By Chebyshev's inequality, we have
$$\begin{array}{l}
3\left(\frac{1}{p(p+1)}+\frac{1}{q(q+1)}+\frac{1}{r(r+1)}\right) \geqslant \\
\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\right)\left(\frac{1}{p+1}+\frac{1}{q+1}+\frac{1}{r+1}\right)
\end{array}$$

That is,
$$\begin{array}{l}
3\left[\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\right)-\left(\frac{1}{p+1}+\frac{1}{q+1}+\frac{1}{r+1}\right)\right] \geqslant \\
\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\right)\left(\frac{1}{p+1}+\frac{1}{q+1}+\frac{1}{r+1}\right)
\end{array}$$

That is,
$$3\left[\frac{1}{f(p, q, r)}-\frac{1}{f(p+1, q+1, r+1)}\right] \geqslant \frac{1}{f(p, q, r)} \cdot \frac{1}{f(p+1, q+1, r+1)}$$

Therefore, $f(p+1, q+1, r+1)-f(p, q, r) \geqslant \frac{1}{3}$, with equality if and only if $p=q=r$.