47. From $\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}+\frac{1}{c^{2}+1}=2$, we get $\frac{a^{2}}{a^{2}+1}+\frac{b^{2}}{b^{2}+1}+\frac{c^{2}}{c^{2}+1}=1$. By the Cauchy-Schwarz inequality, we have
$$\begin{array}{l} {\left[\left(a^{2}+1\right)+\left(b^{2}+1\right)+\left(c^{2}+1\right)\right] \cdot\left(\frac{a^{2}}{a^{2}+1}+\frac{b^{2}}{b^{2}+1}+\frac{c^{2}}{c^{2}+1}\right) \geqslant} \\ (a+b+c)^{2} \end{array}$$

That is, $a^{2}+b^{2}+c^{2}+3 \geqslant(a+b+c)^{2}$, which implies $ab+bc+ca \leqslant \frac{3}{2}$.