11. (15 points) Let $f(x)=x^{2}+b x+c(b, c \in$ $\mathbf{R})$. If for all $x \in \mathbf{R}$, we have $f\left(x+\frac{1}{x}\right) \geqslant 0$, and the maximum value of $f\left(\frac{2 x^{2}+3}{x^{2}+1}\right)$ is 1, find the trajectory of the moving point $P(b, c)$.
11. Since $x$ and $\frac{1}{x}$ have the same sign, we have:
$$
\left|x+\frac{1}{x}\right|=|x|+\left|\frac{1}{x}\right| \geqslant 2 \text {. }
$$

According to the problem, for all real numbers $x$ satisfying $|x| \geqslant 2$, we have $f(x) \geqslant 0$.
(1) When $f(x)=0$ has real roots, the real roots of $f(x)=0$ are within the interval $[-2,2]$. Therefore,
$$
f(x)=x^{2}+b x+c
$$

is an increasing function on $[2,+\infty)$, and
$$
\left\{\begin{array} { l } 
{ f ( - 2 ) \geqslant 0 , } \\
{ f ( 2 ) \geqslant 0 , } \\
{ - 2 \leqslant - \frac { b } { 2 } \leqslant 2 , }
\end{array} \text { i.e., } \quad \left\{\begin{array}{l}
4-2 b+c \geqslant 0, \\
4+2 b+c \geqslant 0, \\
-4 \leqslant b \leqslant 4 .
\end{array}\right.\right.
$$

Also, $\frac{2 x^{2}+3}{x^{2}+1}=2+\frac{1}{x^{2}+1} \in(2,3]$, so the maximum value of $f\left(\frac{2 x^{2}+3}{x^{2}+1}\right)$ is $f(3)=1$, i.e., $9+3 b+c=1$.
Thus, $c=-3 b-8$.
Hence, $\left\{\begin{array}{l}4-2 b-3 b-8 \geqslant 0, \\ 4+2 b-3 b-8 \geqslant 0, \\ -4 \leqslant b \leqslant 4,\end{array}\right.$ i.e., $\left\{\begin{array}{l}b \leqslant-\frac{4}{5}, \\ b \leqslant-4, \\ -4 \leqslant b \leqslant 4 .\end{array}\right.$
Solving this, we get $b=-4, c=4$.
Thus, we obtain the point $P(-4,4)$.
(2) When $f(x)=0$ has no real roots, $\Delta=b^{2}-4 c f(3)$, and $f\left(\frac{2 x^{2}+3}{x^{2}+1}\right)$ has no maximum value (as $x$ approaches 2, $f(x)$ approaches $f(2)$).

Therefore, the necessary and sufficient condition for $f\left(\frac{2 x^{2}+3}{x^{2}+1}\right)$ to have a maximum value is $f(2) \leqslant f(3)$, i.e., $4+2 b+c \leqslant 9+3 b+c$.
So, $b \geqslant-5$.
Also, the maximum value of $f\left(\frac{2 x^{2}+3}{x^{2}+1}\right)$ is $f(3)=1$, i.e., $9+3 b+c=1$, thus, $c=-3 b-8$.

From $\Delta=b^{2}-4 c<0$, we get $b^{2}+12 b+32<0$, i.e., $-8<b<-4$.
So, $3 b+c+8=0(-5 \leqslant b<-4)$.
In summary, the trajectory of the moving point $P(b, c)$ is the line segment with endpoints $A(-5,7)$ and $B(-4,4)$.