16. (2006 Zhejiang Province High School Mathematics Competition Training Test Adapted) If \( f(0) = -\frac{1}{2} \), find the continuous function \( f(t) \) for every real number pair \((x, y)\) that satisfies \( f(x+y) \geqslant f(x) + f(y) + f(xy) + 1 \).
16. $\varphi(x)=f(x)-\frac{1}{2}$, then the inequality $f(x+y) \geqslant f(x)+f(y)+f(x y)+1$ can be transformed into
$$
\begin{array}{l}
\varphi(x+y) \geqslant \varphi(x)+\varphi(y)+\varphi(x y), \text { and } \varphi(0)=f(0)-\frac{1}{2}=0, \\
\varphi(x+y)-\varphi(x) \geqslant \varphi(y)+\varphi(x y) .
\end{array}
$$

Since $f(t)$ is a continuous function, when $y>0$,

taking the limit on both sides, we get
$$
\begin{array}{c}
\lim _{y \rightarrow 0} \frac{\varphi(x+y)-\varphi(x)}{y} \geqslant \lim _{y \rightarrow 0} \frac{\varphi(y)}{y}+\lim _{y \rightarrow 0} \frac{\varphi(x y)}{x y} \cdot x, \\
\varphi^{\prime}(x) \geqslant \varphi^{\prime}(0)+\varphi^{\prime}(0) x .
\end{array}
$$

When $y<0$,
$$
\frac{\varphi(x+y)-\varphi(x)}{y} \leqslant \frac{\varphi(y)}{y}+\frac{\varphi(x y)}{x y} \cdot x,
$$

similarly, we get
$$
\varphi^{\prime}(x) \leqslant \varphi^{\prime}(0)+\varphi^{\prime}(0) x,
$$

so $\varphi^{\prime}(x)=\varphi^{\prime}(0)+\varphi^{\prime}(0) x, \varphi(x)=c+\varphi^{\prime}(0) x+\frac{\varphi^{\prime}(0)}{2} x^{2}$.
By $\varphi(0)=f(0)-\frac{1}{2}=0$, we get $c=0$.
Therefore, the solution to the given function inequality is $f(x)=\varphi(x)+\frac{1}{2}=\frac{1}{2}+\varphi^{\prime}(0) x+\frac{\varphi^{\prime}(0)}{2} x^{2}$.