= Solution
For $x$ in the <nonnegative orthant>, the <positive semidefinite matrix> $P$ satisfies $x^TPx\geq0$. The <nonnegative matrix> $N$ satisfies
$$
x^TNx=\sum_{i,j}N_{ij}x_ix_j\geq0,
$$
since every summand is nonnegative. Adding these inequalities proves
$$
\boxed{x^T(P+N)x\geq0\quad(x\geq0)}.
$$
Consequently $A=P+N$ is a <copositive matrix>. The symmetry of $N$ follows from that of $A$ and $P$; entrywise nonnegativity alone need not imply <positive semidefiniteness>.
This gives the inclusion of the <positive-semidefinite-plus-nonnegative cone> in the <copositive cone>. It is only a sufficient construction; the argument does not claim that every <copositive matrix> has this decomposition in arbitrary <dimension>.
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