Show that $\| x \|^2_2 \| x\|_{2k-2}^{2k-2} \leq n^{1-2/k} \| x \|^{2k}_k$

Question

Let $x$ be in $R^n$. Show that:

$$\| x \|^2_2 \cdot \| x\|_{2k-2}^{2k-2} \leq n^{1-2/k} \cdot \| x \|^{2k}_k~.$$

Answer 1

Ok, I figured out the answer.

We have

$$\| x \|_2 \leq n^{1/2-1/k} \| x \|_k \implies \| x \|_2^2 \leq n^{1-2/k} \| x \|_k^2~.$$

See Relations between p norms for the first inequality. Also,

$$\| x\|_{2k-2} \leq \| x \|_k \implies \| x_{2k-2} \|^{2k-2}_{2k-2} \leq \| x \|_k^{2k-2} ~$$

And we are done.