Five. (13 points) Let the quadratic function
$$
f(x)=x^{2}+b x+c(b, c \in \mathbf{R})
$$

intersect the $x$-axis. If for all $x \in \mathbf{R}$, we have
$$
f\left(x+\frac{1}{x}\right) \geqslant 0 \text {, and } f\left(\frac{2 x^{2}+3}{x^{2}+1}\right) \leqslant 1 .
$$

Find the values of $b$ and $c$.
Five, because $\left|x+\frac{1}{x}\right|=|x|+\left|\frac{1}{x}\right| \geqslant 2$, so for all real numbers $x$ satisfying $|x| \geqslant 2$ we have
$$
f(x) \geqslant 0 \text {. }
$$

Thus, the real roots of $f(x)=x^{2}+b x+c=0$ lie within the interval $[-2,2]$.

Therefore, the quadratic function $f(x)=x^{2}+b x+c(b, c \in \mathbf{R})$ is increasing on the interval $[2,+\infty)$, and
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ f ( 2 ) \geqslant 0 , } \\
{ f ( - 2 ) \geqslant 0 , } \\
{ - 2 \leqslant - \frac { b } { 2 } \leqslant 2 }
\end{array} \Rightarrow \left\{\begin{array}{l}
c \geqslant 2|b|-4, \\
-4 \leqslant b \leqslant 4 .
\end{array}\right.\right. \\
\text { Also, } \frac{2 x^{2}+3}{x^{2}+1}=2+\frac{1}{x^{2}+1} \in(2,3] \text {, then } \\
f(3) \leqslant 1,
\end{array}
$$

which implies $9+3 b+c \leqslant 1 \Rightarrow c \leqslant-3 b-8$.
Hence $\left\{\begin{array}{l}-3 b-8 \geqslant 2|b|-4, \\ -4 \leqslant b \leqslant 4\end{array} \Rightarrow\left\{\begin{array}{l}-3 b \geqslant 2|b|+4, \\ -4 \leqslant b \leqslant 4 .\end{array}\right.\right.$
Therefore, only $b=-4$. In this case, $c=4$.