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.


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.

  • $\begingroup$ the OP has $n^{k-2}$ in its formula, you have $n^{k-1}$ if I understood correctly. $\endgroup$ – mpiktas May 16 '11 at 11:57
  • $\begingroup$ 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. $\endgroup$ – Plop May 16 '11 at 12:00
  • 2
    $\begingroup$ @Raskolnikov thank you for the much prettier formatting $\endgroup$ – Plop May 16 '11 at 12:01
  • $\begingroup$ ok, sorry for bothering. $\endgroup$ – mpiktas May 16 '11 at 12:04

$$ \sum_{i =1}^{n}a_{i}b_{i}\leq\left(\sum_{i=1}^{n}a_{i}^{p}\right)^{\frac{1}{p}}\left(\sum_{i =1}^{n}b_{i}^{q}\right)^{\frac{1}{q}} $$

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!


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