${\cal N}_{13}$ denotes the one between the $1_\text{th}$ and $3_\text{th}$ spin. The equation above can be obtained from the non-degenerate ground state given by: $$\begin{aligned}
|\psi_0\rangle&=&\frac{\sqrt{6}}{6}\left(|\frac{1}{2},0,-\frac{1}{2}\rangle+|-\frac{1}{2},0,\frac{1}{2}\rangle\right)\nonumber\\
&-&\frac{\sqrt{3}}{3}\left(|\frac{1}{2},-1,\frac{1}{2}\rangle+|-\frac{1}{2},1,-\frac{1}{2}\rangle\right).\end{aligned}$$ It is interesting to see that the ground-state entanglement between the spin half and spin 1 in the three-spin case is the same as that in the two-spin case.

Due to the SU(2) symmetry in our system, there are following relations between correlation functions and negativities [@Schliemann] $$\begin{aligned}
{\cal N}_{12}=&-\frac{1}3-\frac{2}3\langle{\bf s}_1\cdot{\bf
S}_1\rangle,\label{N12}\nonumber\\
{\cal N}_{23}=&-\frac{1}3-\frac{2}3\langle{\bf S}_1\cdot{\bf
s}_2\rangle,\nonumber\\
{\cal N}_{31}=&-\frac{1}4-\langle{\bf s}_2\cdot{\bf s}_1\rangle,\end{aligned}$$ where we have removed the max function in the negativity, implying that the negative value of ${\cal N}$ indicates no entanglement. Then, we have the relation between the internal energy and the negativities $$\begin{aligned}
U=&-\frac{5}4-{\cal N}_{13}-\frac{3}2({\cal N}_{12}+{\cal N}_{23})
\nonumber\\
=&-\frac{5}4-{\cal N}_{13}-{3}{\cal N}_{12}. \label{relation}\end{aligned}$$ The second equality follows from the exchange symmetry, namely, the Hamiltonian is invariant when exchanging two spin halves. So, for the three-spin case, the internal energy is related to two negativities.

To apply the above result, we consider the the ground-state properties ($T=0$). The Hamiltonian can be rewritten as $$H=\frac{1}{2}[({\bf s}_1+{\bf S}_1+{\bf s}_2)^2-{\bf s}_1^2-{\bf
S}_1^2-{\bf s}_2^2].$$ Then, by the angular momentum coupling theory, the ground-state energy is obtained as $E_0=-7/4$. 