16. Let the quadratic function $f(x)=a x^{2}+b x+c(a>0$ and $b \neq 0)$.
(1) Given $|f(0)|=|f(1)|=|f(-1)|=1$, find the analytical expression of $f(x)$ and the minimum value of $f(x)$;
(2) Given $|b| \leqslant a,|f(0)| \leqslant 1,|f(-1)| \leqslant 1,|f(1)| \leqslant 1$, prove that when $|x| \leqslant$ 1, $|f(x)| \leqslant \frac{5}{4}$.
16. (1) From $|f(0)|=|f(1)|=|f(-1)|$ we know $|c|=1,|a+b+c|=|a-b+c|=1$, so $(a+b+c)^{2}=(a-b+c)^{2}$, which means $4(a+c) b=0$. Since $b \neq 0$, then $a+c=0, a=-c ; a>0, c<0$, hence $c=-1, a=1, b= \pm 1$. Therefore, $f(x)=x^{2} \pm x+1$.
(2) $|2 b|=|(a+b+c)-(a-b+c)| \leqslant|a+b+c|+|a-b+c| \leqslant 2$, so $|b| \leqslant 1$. Also, $|b| \leqslant a$, so $\left|\frac{b}{a}\right| \leqslant 1,\left|-\frac{b}{2 a}\right| \leqslant \frac{1}{2}<1$. Since $|c|=|f(0)| \leqslant 1, f\left(-\frac{b}{2 a}\right)=$ $\frac{4 a c-b^{2}}{4 a}=c-\frac{b^{2}}{4 a}$, thus $\left|f\left(-\frac{b}{2 a}\right)\right| \leqslant|c|+\left|\frac{b^{2}}{4 a}\right|=|c|+\frac{1}{4}\left|\frac{b}{a}\right||b| \leqslant \frac{5}{4}$. The graph of $f(x)$ opens upwards, and for $|x| \leqslant 1,|f(x)|$ reaches its maximum at $x=1, x=-1$ or $x=-\frac{b}{2 a}$, and $|f(1)| \leqslant 1,|f(-1)| \leqslant 1,\left|f\left(-\frac{b}{2 a}\right)\right| \leqslant \frac{5}{4}$, hence $|f(x)| \leqslant \frac{5}{4}$.