How does this inequality follow from Hölder's inequality?

Assuming $m > 1$ and the $a_{ii}$ are nonnegative, use conjugate exponents $m$ and $m/(m-1)$ in Hölder's inequality to get

$$\sum_{i = 1}^n a_{ii} \le \left(\sum_{i = 1}^n 1^{m/(m-1)}\right)^{(m-1)/m} \left(\sum_{i = 1}^n a_{ii}^m\right)^{1/m} = n^{(m-1)/m}\left(\sum_{i = 1}^n a_{ii}^m\right)^{1/m}$$

The inequality is then obtained by raising to the $m$th power.

Holder's inequality in the discrete form is the following.

Let $a_1$, $a_2$,...,$a_n$, $b_1$, $b_2$,...,$b_n$, $\alpha$ and $\beta$ be positives. Prove that: $$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}$$

It's just homogenization of the Jensen inequality for $f(x)=x^k$, where $k>1$.

Now by Holder we obtain: $$n^{m-1}\sum\limits_{i=1}^na_{ii}^m=\left(\sum\limits_{i=1}^n1\right)^{m-1}\sum\limits_{i=1}^na_{ii}^m\geq\left(\sum\limits_{i=1}^n\left(1^{m-1}a_{ii}^m\right)^{\frac{1}{m}}\right)^m=\left(\sum\limits_{i=1}^na_{ii}\right)^m$$