from generalized inequality, we now that for $p>q$, we have $M\_p(\mathbf{x})\ge M\_q(\mathbf{x})$. now I am curious to know if we can find a constant $\alpha(p,q)$ which is only function of $p,q$ such that $M\_p(\mathbf{x})\le \alpha(p,q) M\_q(\mathbf{x})$. if these exists such constant, how can we find it preferably in closed form?



generalized mean inequality extension
Well, if you accept dependence on the the maximum and minimum components (which are the mean values for plus and minus infinity), Mitrinovic's "Analytic Inequalities" book has such bounds on ratios and differences of the two means, even their weighted versions. E.g.,
$$
\alpha(p,q)\leq \left(\frac{q(C^p-C^q) }{(p-q)(C^q-1) }\right)^{1/p} 
\left(\frac{p(C^q-C^p) }{(q-p)(C^p-1) }\right)^{-1/q}
$$
where $C=M\_{\infty}(\mathbf{x})/M\_{-\infty}(\mathbf{x}),$ thus those max and min values must be bounded away from zero and infinity.


