For some diagonal matrix A = $(a_{ij})$ where the main diagonal are all positive numbers ($a_{ii} > 0$ for $1\le i \le n$), prove:
$$ \textbf{v}^T \textbf{A} \textbf{w} \le \|\textbf{A}^{1/2}\textbf{v}\|_2 \|\textbf{A}^{1/2}\textbf{w}\|_2 $$
We also define the matrix square root of A as $\textbf{A}^{1/2} = (a_{ij}^{1/2})$.
I tried playing around with the Cauchy-Schwarz inequality to prove this, but I think I'm just confusing myself even more. Any solution or suggestions?
What is confusing is that the definition of the square root seems odd at first; but all simplifies once you remember that $A$ is assumed to be diagonal: $$ \mathbf{v}^\top A \mathbf{w} = \sum_{i,j} a_{ij}\mathbf{v}_i \mathbf{w}_j \stackrel{(1)}{=} \sum_{i} a_{ii}\mathbf{v}_i \mathbf{w}_i \stackrel{(2)}{=}\sum_{i} \sqrt{a_{ii}}\mathbf{v}_i \cdot \sqrt{a_{ii}} \mathbf{w}_i \stackrel{(3)}{\leq}\sqrt{\sum_{i} a_{ii}\mathbf{v}^2_i } \cdot \sqrt{\sum_{i} a_{ii}\mathbf{w}^2_i } $$ where (1) uses the fact that $A$ is diagonal, (2) the fact that the entries are nonnegative, and (3) is Cauchy-Schwarz.
Now, you can easily check that $ \| A^{1/2} \mathbf{v}\|_2^2 = \sum_{i} a_{ii}\mathbf{v}^2_i $ and similarly for $ \| A^{1/2} \mathbf{w}\|_2^2$, so the RHS above is exactly $\| A^{1/2} \mathbf{v}\|_2\cdot \| A^{1/2} \mathbf{w}\|_2$.