= Solution
Let $w=(1,2,1,0,0)^T$. Direct multiplication gives $Hw=(0,0,0,2,2)^T$, so $w^THw=0$.
Suppose $H=P+N$, where $P$ is a <positive semidefinite matrix> and $N$ is a symmetric <nonnegative matrix>. Since $w\geq0$, both $w^TPw$ and $w^TNw$ are nonnegative. Their sum is zero, so both vanish. In particular,
$$
0=w^TNw=\sum_{i,j=1}^3 w_iw_jN_{ij}.
$$
Every coefficient $w_iw_j$ in this sum is strictly positive and every $N_{ij}$ is nonnegative, forcing $N_{ij}=0$ for $1\leq i,j\leq3$.
Now apply the same argument to all five cyclic shifts of $w$. The <Horn copositive matrix> is cyclically invariant, so every shifted <vector> also has zero <quadratic form>. It follows that the entries of $N$ vanish on every cyclic block of three consecutive indices. Every pair of indices on a five-cycle lies in such a block, hence $N=0$.
This would imply $P=H$. But the <zero quadratic form of a positive semidefinite matrix> would then force $Hw=0$, contradicting $Hw=(0,0,0,2,2)^T$. Therefore $H$ lies outside the <positive-semidefinite-plus-nonnegative cone>. By the previous equivalence, its quartic form is a <nonnegative polynomial> that is not a <sum of squares polynomial>.
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