All Lie algebras in this post are assumed to be finite-dimensional.
A Lie algebra $L$ is called semisimple if its radical $\operatorname{Rad}(L)$, i.e. the unique maximal solvable ideal of $L$, is $0$.
Suppose that $L$ is a Lie algebra and there exist simple ideals $I_1, \dots, I_n\lhd L$ such that $$L = I_1 \oplus \dots \oplus I_n$$ (vector space direct sum = Lie algebra direct sum). Is it true that $L$ is semisimple?
I tried to argue as follows: as a Lie algebra, each $I_t$ is simple and thus semisimple. It thus suffices to show (by induction) that $$\operatorname{Rad}(L_1\oplus L_2) = \operatorname{Rad}(L_1)\oplus \operatorname{Rad}(L_2)$$ for any Lie algebras $L_1, L_2$. However, I am not able to show this. Is this equality even true?
A duplicate was proposed, which is not relevant...
Following my comment, we can suppose that $L_{1},L_{2}$ (replace $L_{i}$ by $L_{i}/Rad(L_{i})$) are semi-simple. Let $J$ be a solvable ideal of $L_{1}⊕L_{2}$. Then necessarily $J\cap L_{i}={0}$, so the projections $P_{i}:L_{1}⊕L_{2}\mapsto L_{i}$ are injective when restricted $J$. But at the same time $P_{i}(J)$ are homomorphic images of a solvable ideal, hence solvable, so $P_{i}(J)={0}$. This shows that the direct sum of semi-simple lie algebras is semi-simple, and in particular that $Rad(L_{1}⊕L_{2})=Rad(L_{1})⊕Rad(L_{2})$.