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 $a_i,b_i,c_i\geq 0$? 



Here $b$ is irrelevant to the question. 
Would anybody explain how Schwarz and Hölder are used in the last inequality? 

[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
$$
\sum a_ib_ic_i\leq \left(\sum a_i^2\right)^{1/2}\left(\sum b_i^4\right)^{1/4}\left(\sum c_i^4\right)^{1/4}.
$$
 A: 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*}
A: 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 $(*)$. 
