Hölder Inequality I am wondering how I get 
$$
 \frac{a_{1}^{k}}{b_{1}}+\frac{a_{2}^{k}}{b_{2}}+\cdots+\frac{a_{n}^{k}}{b_{n}}\geq\frac{\left(a_{1}+\cdots+a_{n}\right)^{k}}{n^{k-2}\cdot\left(b_{1}+\cdots+b_{n}\right)}. 
$$
from the Hölder inequality
$$ \sum_{i =1}^{n}a_{i}b_{i}\leq\left(\sum_{i=1}^{n}a_{i}^{p}\right)^{\frac{1}{p}}\left(\sum_{i =1}^{n}b_{i}^{q}\right)^{\frac{1}{q}}. $$
I was reading through AoPS and I am struggling to see how the first was obtained from the second.
 A: First we apply 
$$\sum_{i =1}^{n}x_{i}y_{i}\leq\left(\sum_{i=1}^{n}x_{i}^{p}\right)^{\frac{1}{p}}\left(\sum_{i =1}^{n}y_{i}^{q}\right)^{\frac{1}{q}}$$ 
with $p=k$, $q=k/(k-1)$, $x_i=a_i/b_i^{1/k}$, $y_i=b_i^{1/k}$ ,
to get 
$$\sum_i a_i \leq \left(a_i^k/b_i \right)^{1/k} \left( b_i^{1/(k-1)} \right)^{(k-1)/k} \; ,$$
which is equivalent to 
$$\left( \sum_i a_i \right)^k \leq \left( \sum_i a_i^k/b_i \right) \left( \sum_i b_i^{1/(k-1)} \right)^{k-1} \; .$$
By concavity of $x \mapsto x^{1/(k-1)}$ (I guess $k \geq 2$), we also have that 
$$1/n \sum_i b_i^{1/(k-1)} \leq \left( \sum_i b_i/n \right)^{1/(k-1)}$$
which combined with the preceding inequality, gives the desired result.
A: $$ \sum_{i =1}^{n}a_{i}b_{i}\leq\left(\sum_{i=1}^{n}a_{i}^{p}\right)^{\frac{1}{p}}\left(\sum_{i =1}^{n}b_{i}^{q}\right)^{\frac{1}{q}} $$
I think, this form of the Holder's inequality is not conveniently for an inequalities proofs.
It's better to write Holder in the following form.
Let $a_i>0$, $b_i>0$, $\alpha>0$ and $\beta>0$. Prove that:
$$(a_1+a_2+...+a_n)^{\alpha}(b_1+b_2+...+b_n)^{\beta}\geq\left(\left(a_1^{\alpha}b_1^{\beta}\right)^{\frac{1}{\alpha+\beta}}+\left(a_2^{\alpha}b_2^{\beta}\right)^{\frac{1}{\alpha+\beta}}+...+\left(a_n^{\alpha}b_n^{\beta}\right)^{\frac{1}{\alpha+\beta}}\right)^{\alpha+\beta}$$
By the last inequality we obtain:
$$
 n^{k-2}\left(b_{1}+\cdots+b_{n}\right)\left(\frac{a_{1}^{k}}{b_{1}}+\frac{a_{2}^{k}}{b_{2}}+\cdots+\frac{a_{n}^{k}}{b_{n}}\right)=$$
$$=(1+1+...+1)^{k-2}\left(b_{1}+\cdots+b_{n}\right)\left(\frac{a_{1}^{k}}{b_{1}}+\frac{a_{2}^{k}}{b_{2}}+\cdots+\frac{a_{n}^{k}}{b_{n}}\right)\geq\left(a_{1}+\cdots+a_{n}\right)^{k} 
$$
and we are done!
