how does one prove $\operatorname{tr}(A^2(A^2)^T) \le\operatorname{tr}(AA^T)^2$?

Question

I would like to prove that $\lim_{H \to 0_{k\times k}} \frac {H^2}{\vert {H} \vert} = 0_{k\times k}$.

here is my idea:

since $M_{k\times k}(\Bbb R)$ is an inner product space with respect to the standard inner product defined by $(A,B)=\operatorname{tr}(AB^T)$, the norm defined by the standard inner product is $\vert A \vert =\sqrt{(A,A)}$ hence if $\operatorname{tr}(A^2(A^2)^T) \le \operatorname{tr}(AA^T)^2$ is true for every matrix $A \in M_{k\times k}(\Bbb R)$ then: $$0 \le \lim_{H \to 0_{k\times k}} \vert \frac {H^2}{\vert {H} \vert}\vert = \lim_{H \to 0_{k\times k}} \frac {1} {\vert H \vert} \vert H^2 \vert = \lim_{H \to 0_{k\times k}} \frac {1} {\vert H \vert} \operatorname{tr}(H^2(H^2)^T)$$

$$\le \lim_{H \to 0_{k\times k}} \frac {1} {\vert H \vert} \vert H \vert^2=\lim_{H \to 0_{k\times k}} |H|=0$$ hence by the squeeze theorem we get $\lim_{H \to 0_{k\times k}} |\frac {H^2}{\vert {H} \vert}| = 0$ and this can only occur if $\lim_{H \to 0_{k\times k}} \frac {H^2}{\vert {H} \vert} = 0_{k\times k}$.

Answer 1

$(i,j)$-entry of $H^2$ where $H=(h_{ij})$ is $$t=\sum_k\ h_{ik}h_{kj} = v\cdot w$$ where $v=(h_{i1},\cdots , h_{in}),\ w= (h_{1j},\cdots, h_{nj})$ Then $|v|\leq |H|,\ |w|\leq |H|$. Hence $$ |v\cdot w|\leq |v||w| \leq |H|^2 $$

So $ \lim\ |\frac{t}{|H|}| \leq \lim\ \frac{|H|^2}{|H|} = \lim\ |H| =0 $

Answer 2

$\text{trace}\Big(A^2(A^2)^*\Big)$
$=\text{trace}\Big((A^*A)^*(AA^*) \Big)$
$\leq \text{trace}\Big((A^*A)^2\Big)^\frac{1}{2} \cdot \text{trace}\Big((AA^*)^2\Big)^\frac{1}{2}$
$=\text{trace}\Big((AA^*)^2\Big)$

by Cauchy-Schwarz, with equality iff $A$ is normal.