= Solution
Let $r=|\boldsymbol r_1-\boldsymbol r_2|$ and $\widehat{\boldsymbol r}=(\boldsymbol r_1-\boldsymbol r_2)/r$. A differentiable central <potential energy> gives the internal <forces>
$$
\boldsymbol F_{12}=-\nabla_{\boldsymbol r_1}U=-U'(r)\widehat{\boldsymbol r},\qquad
\boldsymbol F_{21}=-\nabla_{\boldsymbol r_2}U=+U'(r)\widehat{\boldsymbol r}.
$$
There are no external <forces>. Differentiating the total <kinetic energy> and using <Newton's second law> therefore gives
$$
\begin{aligned}
\frac{dT}{dt}
&=m_1\boldsymbol v_1\cdot\dot{\boldsymbol v}_1
+m_2\boldsymbol v_2\cdot\dot{\boldsymbol v}_2\\
&=\boldsymbol F_{12}\cdot\boldsymbol v_1+
\boldsymbol F_{21}\cdot\boldsymbol v_2\\
&=-U'(r)\widehat{\boldsymbol r}\cdot(\boldsymbol v_1-\boldsymbol v_2).
\end{aligned}
$$
Since $\dot r=\widehat{\boldsymbol r}\cdot(\boldsymbol v_1-\boldsymbol v_2)$, the last expression is $-U'(r)\dot r=-dU/dt$. Thus
$$
\boxed{\frac{dT}{dt}+\frac{dU}{dt}=0,\qquad T+U=\mathrm{constant}.}
$$
This conserves total <mechanical energy>, including the <kinetic energy> of centre-of-mass motion, not merely the relative <kinetic energy>.
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