4.32 If the algebraic expression is set as
$$
P=\frac{a^{2}}{a^{2}+b^{2}+c^{2}}+\frac{b^{2}}{b^{2}+c^{2}+d^{2}}+\frac{c^{2}}{c^{2}+d^{2}+a^{2}}+\frac{d^{2}}{d^{2}+a^{2}+b^{2}} \text {. }
$$

where $a, b, c, d$ are all non-zero real numbers. Then
(A) $0<P \leqslant 1$.
(B) $1<P<2$.
(C) $2 \leqslant P<3$.
(D) $P \geqslant 3$.
(China Beijing High School Mathematics Competition, 1986)
[Solution] $P>\frac{a^{2}}{a^{2}+b^{2}+c^{2}+d^{2}}+\frac{b^{2}}{a^{2}+b^{2}+c^{2}+d^{2}}$
$$
\begin{aligned}
& +\frac{c^{2}}{a^{2}+b^{2}+c^{2}+d^{2}}+\frac{d^{2}}{a^{2}+b^{2}+c^{2}+d^{2}} \\
= & \frac{a^{2}+b^{2}+c^{2}+d^{2}}{a^{2}+b^{2}+c^{2}+d^{2}}=1 . \\
\text { Also, } \quad P< & \frac{a^{2}}{a^{2}+c^{2}}+\frac{b^{2}}{b^{2}+d^{2}}+\frac{c^{2}}{a^{2}+c^{2}}+\frac{d^{2}}{b^{2}+d^{2}} \\
= & \left(\frac{a^{2}}{a^{2}+c^{2}}+\frac{c^{2}}{a^{2}+c^{2}}\right)+\left(\frac{b^{2}}{b^{2}+d^{2}}+\frac{d^{2}}{b^{2}+d^{2}}\right) \\
& =2 . \\
\therefore \quad 1< & P<2 .
\end{aligned}
$$

Therefore, the correct choice is $(B)$.