How to prove $ \sum a_ib_ic_i\leq \left(\sum a_i^2\right)^{1/2}\left(\sum b_i^2\right)^{1/2}\left(\sum c_i^2\right)^{1/2} $ with $a_i,b_i,c_i\geq 0$?

Question

Here $b$ is irrelevant to the question.

Would anybody explain how Schwarz and Hölder are used in the last inequality?

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[Added:] Things boil down to proving $$ \sum a_ib_ic_i\leq \left(\sum a_i^2\right)^{1/2}\left(\sum b_i^2\right)^{1/2}\left(\sum c_i^2\right)^{1/2} $$ where $a_i,b_i,c_i\geq 0$. But I only got $$ \sum a_ib_ic_i\leq \left(\sum a_i^2\right)^{1/2}\left(\sum b_i^4\right)^{1/4}\left(\sum c_i^4\right)^{1/4}. $$

Answer 1

As said in my comment:

first do Cauchy-Schwartz on the $i$ index, then on the $j$ index (or conversely). Then it follows directly \begin{align*}\sum_{ij}a_ib_{ij}c_{j}& =\sum_j\left(\sum_ia_ib_{ij}\right)c_j\\ & \leq\left(\sum_ia_i^2\right)^{1/2}\cdot \sum_j\left(\sum_ib_{ij}^2\right)^{1/2}\cdot c_j\\ & \leq \left(\sum_ia_i^2\right)^{1/2}\cdot \left(\sum_j\underbrace{\left(\left(\sum_ib_{ij}^2\right)^{1/2}\right)^{2}}_{=\sum_ib_{ij}^2}\right)^{1/2}\cdot \left(\sum_jc_j^2\right)^{1/2}\\ & =\left(\sum_ia_i^2\right)^{1/2}\cdot \left(\sum_{ji}b_{ij}^2\right)^{1/2}\cdot \left(\sum_jc_j^2\right)^{1/2} \end{align*}

Answer 2

Thanks to all the comments and answer, I could have used matrices. Essentially, $$ |(a,Bc)|\leq |a||Bc| $$ where $a,c\in\mathbb{R}^2$ and $B$ is a $2\times 2$ matrix. Now all one needs is $$ |Bc|\leq \|B\|_F\cdot|c|\tag{*} $$ where $\|\|_F$ is the Frobenius norm for matrices. If one is satisfied with a rough bound, then $$ |Bc|\leq C\|B\|_F\cdot|c| $$ is immediate since we are working in a finite dimensional space. Otherwise, doing it component-wise (as the accepted answer does) for the $|Bc|$ part gives the exact bound $(*)$.