The set of ideals of a solvable lie algebra L is a chain?

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I know that every finite dimensional lie algebra over a field $\mathcal{F}$ has a unique maximal solvable ideal, all subalgebras of a solvable lie algebra are also solvable, and a sum of solvable lie algebras is solvable.

For a solvable lie algebra L, its maximal ideal is unique. By induction, the set of ideals of L is a chain relative to the order $\subset$. The maximal ideal of L is a sum of all proper ideals. I want to know if the maximal ideal is [LL]. Let $L^{(1)}=[LL]$, $L^{(i)}=[L^{(1-1)}L^{(1-1)}]$. Suppose that $L^{(n)}=0$, $n \in \mathbb{Z}_{+}$.What is the connection between
\begin{align*} 0 \subset [L^{n-1}L^{n-1}] \subset \ldots \subset [L^{1}L^{1}] \subset [LL] \subset L \end{align*} and the chain of ideals of $L$.

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This is not true, suppose $L$ is commutative of dimension 2, every 1-dimensional subspace is a maximal ideal, so there is not a unique maximal ideal and $[L,L]=0$.