Let $L$ be a Lie algebra and $\mathrm{Rad}(L)$ its unique maximal solvable ideal.
Problem: Show that if $B$ is a maximal solvable subalgebra of $L$ (i.e. a Borel subalgebra) then $\mathrm{Rad}(L)\subseteq B$.
I am not sure why. We know that $\mathrm{Rad}(L)$ is a solvable subalgebra, so it is contained in at least one maximal solvable subalgebra. But why all of them?