I need to prove the following inequality:
$$\|x\|_{p} \geqslant \|x\|_{q}$$ if $q \geqslant p$.
I tried to do it using this inequality:
$$\frac{d\left(\sum_{i=1}^{d}|x|^{\alpha}\right)^{\frac{1}{\alpha}} }{d \alpha} \leqslant 0$$
After differentiation I got stuck in the following equation:
$$\left(\left(\sum_{i=1}^{d}|x_i|^{\alpha})^{\frac{1}{\alpha}}\right) \ln\left(\sum_{i=1}^{d}|x_i|^{\alpha}\right) \left(\sum_{i=1}^{d}(|x_i|^{\alpha} \ln(|x_i|\right) \right)$$
Could someone help me to prove it, please?
Obviously when $p=q$, $\|x\|_p=\|x\|_q$. Generally it should be $$ p<q\implies \|x\|_p<C\|x\|_q\tag 1 $$
By Holder's inequality $$ \|x\|_p^p=\sum_{i=1}^d|x_i|^p\leqslant \left(\sum_{i=1}^d(|x_i|^p)^{\frac{q}{p}}\right)^{\frac{p}{q}}\left(\sum_{i=1}^d1^{\frac{q}{q-p}}\right)^{\frac{q-p}{q}}=\left(\sum_{i=1}^d(|x_i|^q)\right)^{\frac{p}{q}}C^p=\|x\|_q^pC^p $$ Where $C=d^{\frac{q-p}{pq}}$. So $(1)$ holds.