Let the $p$-norm $\|x\|_p$ of $x\in\mathbb{R}^N$ be defined as $\left(\sum_{n=1}^N |x_n|^p\right)^{1/p}$. Notice then that the left hand side of the inequality is $(\|x\|_{2(p-1)})^{p-1}$, and the right hand side is $(\|x\|_p)^{p-1}$. Since we take $p\gt 1$, this implies that we can remove the powers and the inequality will be the same inequality as the one for $\|x\|_{2(p-1)}$ and $\|x\|_p$. Now since $p\gt 1$, there is a fact that the higher order p-norm is lesser than or equal to the lower order $p$-norm (intuition: higher order $p$-norms define a “stricter” sense of distance). Proof: let $b\gt a\gt 1$. Now, think about the expansion of the multinomial, when all values in the brackets are positive: $(\|x\|_a)^b\ge\sum_{n=1}^N |x_n|^{a\cdot (b/a)}=(\|x\|_b)^b$ which shows that the $a$-norm of $x$, the lower order norm, was greater than or equal to than the $b$-norm of $x$ - key here is the fact that both $a,b\gt 1$. This proves your inequality for all $p\ge2$ since $2(p-1)\ge p$ if $p\ge2$, so the $p$-norm is a lower order norm than the $2(p-1)$-norm, therefore the $p$-norm of $x$ is greater than or equal to, therefore the right hand side is greater than or equal to the left hand side.
It is false for $p$ in the range $(1,2)$ since in those ranges $p\gt 2(p-1)$, and therefore the norm in the LHS is of lower order, not higher, making the left hand side larger.
For example, take $x=(0.5,0.5)\in\mathbb{R}^2$, with $p=1.5$. The LHS is $(0.5^1+0.5^1)^{1/2}=1$, but the RHS is $(0.5^{3/2}+0.5^{3/2})^{1/3}=0.89089871813\lt1\lt LHS$.