TITLE: IMO $2001$ problem $2$
QUESTION [5 upvotes]: Let $a,b,c \in \mathbb{R}$. Prove that $$\frac{a}{\sqrt{a^2+8bc}} + \frac{b}{\sqrt{b^2+8ca}}+ \frac{c}{\sqrt{c^2+8ab}} \geqslant 1.$$

I tried to follow the proposed solution for this which depended on Hölder's inequality, but I'm a bit confused about how they came up with the expression. How I remember Hölder's is that it states that $$\sum_{i=1}^n |x_iy_i| \leqslant (\sum_{i=1}^n|x_i|^p)^{1/p}(\sum_{i=1}^n|y_i|^q)^{1/q}$$
and we need the same Conjugate property as in Young's inequality $\frac{1}{p} + \frac{1}{q} =1.$
What they had was $$(\sum \frac{a}{\sqrt{a^2+8bc}})(\sum \frac{a}{\sqrt{a^2+8bc}})(\sum a(a^2+8bc)) \geqslant (a+b+c)^3.$$
From here it was quite straightforward, but any clarification on how we can get this result from Hölder's would be appreciated.

REPLY [2 votes]: The Holder's inequality for two sequences it's the following.

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

For positives $a$, $b$ and $c$ by Holder we obtain:
$$\left(\sum_{cyc}\frac{a}{\sqrt{a^2+8bc}}\right)^2\sum_{cyc}a(a^2+8bc)\geq$$
$$\geq\left(\sum_{cyc}\sqrt[3]{\left(\frac{a}{\sqrt{a^2+8bc}}\right)^2a(a^2+8bc)}\right)^3=(a+b+c)^3.$$
In our case $n=3$, $\alpha=2$ and $\beta=1$.