= Solution
An <Ideal of a Lie algebra> is a <vector subspace> $I\subseteq\mathfrak g$ such that $[\mathfrak g,I]\subseteq I$. The <derived series of a Lie algebra> is defined by $\mathfrak g^{(0)}=\mathfrak g$ and $\mathfrak g^{(r+1)}=[\mathfrak g^{(r)},\mathfrak g^{(r)}]$.
Suppose that $I$ is an ideal and $x\in\mathfrak g$, $u,v\in I$. The <Jacobi identity> gives
$$
[x,[u,v]]=[[x,u],v]+[u,[x,v]].
$$
Both terms on the right lie in $[I,I]$, because $[x,u],[x,v]\in I$. Thus $[I,I]$ is again an ideal. Starting from the ideal $\mathfrak g$ and applying this observation inductively proves that \b[every term $\mathfrak g^{(r)}$ of the derived series is an ideal of $\mathfrak g$].
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