Let $\mathfrak{g}$ be a Lie algebra over algebraically closed field $k$ of characteristic $0$.
The radical $R(\mathfrak{g})$ is the largest solvable ideal of $\mathfrak{g}$.
The Killing form $\kappa$ is the trace form of the adjoint representation of $\mathfrak{g}$.
The radical of the Killing form on $\mathfrak{g}$ is $\mathfrak{g}^\perp = \{x \in \mathfrak{g} : \kappa(x,y) = 0, \forall y \in \mathfrak{g} \}.$
Clearly it is true that $\mathfrak{g}^\perp \subset R(\mathfrak{g})$, as by the invariance of the Klling form $\mathfrak{g}^\perp$ is an ideal, and so restricting $\kappa$ to $\mathfrak{g}^\perp$ we find that $\kappa$ is trivial and so $\mathfrak{g}^\perp$ is solvable by Cartan's criterion.
I thought I had proved also $R(g) \subset \mathfrak{g}^\perp$ but this does not appear to be true.
Where does it appear that I am making an error in the below attempted proof?
Since $ \mathfrak{g}^\perp$ ideal by invariance, consider $\mathfrak{g} / \mathfrak{g}^\perp$. Since $\mathfrak{g} / \mathfrak{g}^\perp$ has non-degenerate Killing form (as we have quotiented by its kernel), by the Cartan criterion $\mathfrak{g} / \mathfrak{g}^\perp$ is semisimple, and so $R(\mathfrak{g}) \subset \mathfrak{g}^\perp$.
Like I mentioned in my comment, the Killing form of the quotient Lie algebra $\mathfrak{g}/\mathfrak{g}^\perp$ does not necessarily equal the reduction of the Killing form of $\mathfrak{g}$ to $\mathfrak{g}/\mathfrak{g}^\perp$.
Elaborating: As mentioned in the comment of Torsten Schoeneberger here, a good example is a 3-dimensional Lie algebra so that the corresponding Killing form has a two-dimensional radical like here (whenever $\lambda^2 \neq -1$). The Lie algebra $\mathfrak{g}/\mathfrak{g}^\perp$ must be one-dimensional, thus abelian, thus the canonical Killing form on $\mathfrak{g}/\mathfrak{g}^\perp$ vanishes. However, the reduction of the Killing form of $\mathfrak{g}$ does not vanish on the one-dimensional quotient, as we constructed it to be non-degenerate.