**No.** A Lie algebra satisfying that property is called perfect. For an example of a perfect Lie algebra that isn't semisimple, take a semisimple $L$ and an irreducible representation $V$ of $L$, and define a bracket on $L \times V$ by
$$ [(X,v),(Y,u)] := ([X,Y],Xu-Yv). $$
This turns $L \times V$ into a perfect Lie algebra with $\text{Rad}(L \times V) = V$.