5. Given $f(x)=x|x|$. If for any $x \geqslant 1$ there is $f(x+m)+m f(x)<0$, then the range of real number $m$ is ( ).
(A) $(-\infty,-1)$
(B) $(-\infty,-1]$
(C) $(-\infty,-2)$
(D) $(-\infty,-2]$
5. B.

Obviously, $m<0$, and $-m f(x)=f(\sqrt{-m} x)$.
Also, $f(x)=x|x|$ is an increasing function and an odd function, so
$$
\begin{array}{l}
f(x+m)+m f(x)<0 \\
\Leftrightarrow f(x+m)<-m f(x) \\
\Leftrightarrow f(x+m)<f(\sqrt{-m} x) \\
\Leftrightarrow x+m<\sqrt{-m} x .
\end{array}
$$