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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.

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up vote 5 down vote accepted

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.

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the OP has $n^{k-2}$ in its formula, you have $n^{k-1}$ if I understood correctly. – mpiktas May 16 '11 at 11:57
In the last inequality, there is $1/n$ on the left and $1/n^{1/(k-1)}$ on the right, so if we take the $k-1$th power of this inequality, we get $1/n^{k-1}$ on the left and $1/n$ on the right. – Plop May 16 '11 at 12:00
@Raskolnikov thank you for the much prettier formatting – Plop May 16 '11 at 12:01
ok, sorry for bothering. – mpiktas May 16 '11 at 12:04

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