$\left(\sum_{k=1}^{n}a_k^q\right)^{\frac{1}{q}}\leq \left(\sum_{i=1}^{n}a_o^i\right)^{\frac{1}{p}}$ [duplicate]

Question: Let $1\leq p<q$ real numbers, $n$ be a positive integer, and $a_1,a_2,\dots,a_n$ be nonnegative real numbers. Then prove that $$\left(\sum_{k=1}^{n}a_k^q\right)^{\frac{1}{q}}\leq \left(\sum_{i=1}^{n}a_i^p\right)^{\frac{1}{p}}$$

If $q>1$, $(a+b)^q\geq a^q+b^q$. Therefore, $\left(\sum_{k=1}^{n}a_k^q\right)\leq \left(\sum_{i=1}^{n}a_i\right)^{q}$. In other words, if $p=1$, then the inequality is easy. But how can I prove the inequality if $p>1$?

Edit: I have tried using a generalized mean, but I am unable to eliminate $w_i$, the weights involved in the function

Replace $a_i^p$ by $b_i$, denote $q/p=r>1$. Then your inequality becomes

$$\sum b_i\ge \left(\sum b_i^r\right)^{1/r}.$$

You wrote in the OP that this inequality is easy (and it is).