Dimension of the nilradical nil( g ) of a finite-dimensional Lie algebra g

Question

The nilradical nil(g) of a finite-dimensional Lie algebra g is its maximal nilpotent ideal, which exists because the sum of any two nilpotent ideals is nilpotent. How to prove that $dim(nil(g))$ at least $1/2 dim(g)$?

Answer 1

The correct statement is, the the nilradical of a solvable Lie algebra $L$ over an algebraically closed field of characteristic zero satisfies $$ \dim {\rm nil}(L)\ge \frac{1}{2}\dim (L). $$ Actually, more is true, we have $$ \dim {\rm nil}(L)\ge \frac{1}{2}(\dim (L)+\dim (L^{(2)})) $$

Reference: Snobl, $(18)$, page $8$.