Prove that all ternary trees (each vertex has up to three children) with $n$ vertices have a height of at least $\lfloor\log_{3}{n}\rfloor$.
I know we can show that a perfect ternary tree of height $h$ has $3^0+3^1+3^2+\dots+3^h=\frac{3^{h+1}-1}{2}$ vertices, $$n=\frac{3^{h+1}-1}{2}$$ so a perfect ternary tree with n vertices must have a height of $$h=\log_{3}{(2n+1)}-1$$ So, imperfect ternary trees have a height of at most $h=\log_{3}{(2n+1)-1}$, and $\lfloor\log_{3}{n}\rfloor\leq \log_{3}{(2n+1)}$, but I need to give a lower bound of $\lfloor\log_{3}{n}\rfloor$.
How should I continue? Or is there a better way to prove this?