Q:

Show the Trace$(B)^{2} \leq$ nTrace$(B^{T}B)$

The following definition is needed for my actual question:
For $A, B \in \mathcal{M}_{n \times n}$ define
$$
 \langle A, B \rangle = \text{Trace}(B^{T}A) = \sum_{j=1}^{n}\sum_{i=1}^{n}b_{ij} \, a_{ij}
$$
And here is the actual question:
For $B \in \mathcal{M}_{n \times n}$ show that
$$
\text{Trace}(B)^{2} \:\leq\: n \,\text{Trace}(B^{T}B)
$$
I know I have to use the Cauchy-Schwarz Inequality, but I wasn't exactly sure how to.  Here is my attempt:
Since I only have one matrix I decided to use the identity matrix
$$
\begin{align}
\text{Trace}(B)^{2} \:&\leq\: n \,\text{Trace}(B^{T}B) \\
\text{Trace}(I^{T}B)^{2} \:&\leq\: n \, \langle B, B \rangle \\
\langle I_{n}, B \rangle^{2} \:&\leq\: n \, \Vert B \Vert^{2} \\
\langle I_{n}, B \rangle \:&\leq\: \sqrt{n} \, \Vert B \Vert
\end{align}
$$
I feel like I either did it incorrect, or that I'm close but am missing something.

A:

It seems like you pretty much have the solution:
$$
\operatorname{trace}(B)^2 = 
\overbrace{\langle I,B \rangle^2 \leq
\langle I,I \rangle \langle B,B \rangle}^{\text{Cauchy Schwarz}}  =
\operatorname{trace}(I) \operatorname{trace}(B^TB) = 
n\operatorname{trace}(B^TB)
$$
So, we have
$$
\operatorname{trace}(B)^2 \leq n\operatorname{trace}(B^TB)
$$
which is what we wanted.

