Suppose $1 < p \leq q \leq r$ and $x \in \mathcal {R}^n $, by Liapunov's Inequality, if for $\lambda \in (0,1)$ and $q=\lambda p+(1-\lambda)q$, then
$$ \lVert x\rVert_q^q \leq \lVert x\rVert_p^{\lambda p}\lVert x\rVert_r^{(1-\lambda)r}. $$
I am wondering if there is an inverse inequality
$$ \lVert x\rVert_q^q \geq \lVert x\rVert_p^{\lambda p}\lVert x\rVert_r^{(1-\lambda)r}n^{a} $$ where $a$ is some constant.
It's obvious that $\lVert x\rVert_q \geq \lVert x\rVert_r$, and $\lVert x\rVert_q \geq n^{1/q-1/p} \lVert x\rVert_p$, but it's not tight. So is there a tight inequality like Liapunov's Inequality?