14. Given the function $f(x)=\frac{2 x^{2}+b x+c}{x^{2}+1}(b<0)$ has a range of $[1,3]$.
(1) Find the values of $b$ and $c$;
(2) Determine the monotonicity of the function $F(x)=\lg f(x)$ when $x \in[-1,1]$, and prove your conclusion;
(3) If $t \in \mathbf{R}$, prove that $\lg \frac{7}{5} \leqslant F\left(\left|t-\frac{1}{6}\right|-\left|t+\frac{1}{6}\right|\right) \leqslant \lg \frac{13}{5}$.
14. Solution (1) Let $y=\frac{2 x^{2}+b x+c}{x^{2}+1}$, then $(y-2) x^{2}-b x+y-c=0$.

Since $x \in \mathbf{R}$, the discriminant of (1) $\Delta \geqslant 0$, i.e., $b^{2}-4(y-2)(y-c) \geqslant 0$, which means
$$
4 y^{2}-4(2+c) y+8 c-b^{2} \leqslant 0 .
$$

By the given condition, the solution set of inequality (2) is $[1,3]$,
so 1 and 3 are the roots of the equation $4 y^{2}-4(2+c) y+8 c-b^{2}=0$.
$$
\left\{\begin{array}{l}
1+3=2+c \\
1 \times 3=\frac{8 c-b^{2}}{4}
\end{array},\right.
$$

Thus, $c=2, b=-2, b=2$ (both).
(2) For any $x_{1}, x_{2} \in[-1,1]$, and $x_{2}>x_{1}$, then $x_{2}-x_{1}>0$, and $\left(x_{2}-x_{1}\right)\left(1-x_{1} x_{2}\right)>0$,
so $f\left(x_{2}\right)-f\left(x_{1}\right)=-\frac{2 x_{2}}{1+x_{2}^{2}}-\left(-\frac{2 x_{1}}{1+x_{1}^{2}}\right)=\frac{2\left(x_{2}-x_{1}\right)\left(1-x_{1} x_{2}\right)}{\left(1+x_{1}^{2}\right)\left(1+x_{2}^{2}\right)}>0$,
so $f\left(x_{2}\right)>f\left(x_{1}\right), \lg f\left(x_{2}\right)>\lg f\left(x_{1}\right)$, i.e., $F\left(x_{2}\right)>F\left(x_{1}\right)$.
Therefore, $F(x)$ is an increasing function.
(3) Let $u=\left|t-\frac{1}{6}\right|-\left|t+\frac{1}{6}\right|,|u| \leqslant\left|\left(t-\frac{1}{6}\right)-\left(t+\frac{1}{6}\right)\right|=\frac{1}{3} \cdot$,

i.e., $-\frac{1}{3} \leqslant u \leqslant \frac{1}{3}$. According to the monotonicity of $F(x)$, $F\left(-\frac{1}{3}\right) \div F(u) \leqslant F\left(\frac{1}{3}\right)$,
so $\lg \frac{7}{5} \leqslant F\left(\left|t-\frac{1}{6}\right|-\left|t+\frac{1}{6}\right|\right) \leqslant \lg \frac{13}{5}$ holds for any real number $t$.