Show that $\left(\sum_{n=1}^{\infty} a_n^q \right)^{1/q}$ is a decreasing function of $q$ for $q>0$ 
Let $\{a_n\} $ be a sequence of non-negative real number such that $\sum_{n=1}^{\infty} a_n^q$ is convergent . Then show that $\left(\sum_{n=1}^{\infty} a_n^q \right)^{1/q}$ is a decreasing function of $q$ for $q>0$


Since given that $\sum_{n = 1}^{\infty} a_n^q$ is convergent  therefore $\sum_{n=1}^{\infty} a_n^q=c, c>0$ then   $\left(\sum_{n = 1}^{\infty} a_n^q \right)^{1/q}=c^{1/q}$ .
Now if $c>1 $ then $ c^{1/q} $ is decreasing function of $q$ but if $0<c<1 $ then $c^{1/q}$ is increasing function of $ q $.

As D F said in the comment,
$$\sum_{j = 1}^\infty a_j^q = c(q),$$
so your argument doesn't work. For $p < q$, we have
$$\frac{|a_j|}{\|a\|_{q}} \le 1, \quad \text{for }~\|a\|_{q} = \left(\sum_{j = 1}^\infty a_j^q\right)^{1/q}.$$
Therefore,
$$\left|\frac{a_j}{\|a\|_q}\right|^q \le \left|\frac{a_j}{\|a\|_q}\right|^p,$$
and you conclude the statement by summation.
