trace of symmetric matrix problems I have the two problems below from a practice exam. I can prove them on their own but am not exactly sure if/how to show that they only hold for symmetric matrices and for '3)' showing that it only holds for a matrix with only positive eigenvalues. I know that if the eigenvalues are all positive the determinant will be positive and the trace but cant see how that affects whether '3)' is true or not.


*Show that $\operatorname{Tr}(A^2) \leq \operatorname{Tr}(A)^2$ holds for any symmetric matrix $A$ whose eigenvalues are all non-negative.

*Show that $\operatorname{Tr}(AB)^2 \le \operatorname{Tr}(A^2)\operatorname{Tr}(B^2)$ holds for any symmetric matrices $A$ and $B$.


About question  $\#4$:
Notation: Let $C= AB$ and $K^2_{ii}$ denote the element at the $(i,i)$ position of the matrix $K^2$. 
Firstly, due to symmetry of the $n\times n$ matrices $A,B$, it is easy to prove that:
$$c_{ii} \le \left(A^2_{ii}\right)^{1/2} \cdot \left(B^2_{ii}\right)^{1/2},\quad i = 1,\ldots, n.\tag 1$$

Proof of $(1)$
$c_{ii} =^\color{red}{\star\star} a_{i1} \cdot b_{i1} + \cdots +a_{in}\cdot b_{in}= \sum\limits_{j=1}^{n}a_{ij}\cdot b_{ij}\color{blue}{\le^\star} \left(\sum\limits_{j=1}^{n}a^2_{ij}\right)^{1/2}\cdot \left(\sum\limits_{j=1}^nb_{ij}^2\right)^{1/2}=\left(A^2_{ii}\right)^{1/2}\cdot \left(B^2_{ii}\right)^{1/2}\quad \text{QED}$
Thus, we have:
$$\big[\operatorname{trace} (AB)\big]^2=\left(\sum_{i=1}^n c_{ii}\right)^2\le\left[ \sum_{i=1}^n \left(A^2_{ii}\right)^{1/2}\cdot \left(B^2_{ii}\right)^{1/2}\right]^2\color{blue}{\le^\star}\sum_{i=1}^n A^2_{ii} \cdot \sum_{i=1}^n B^2_{ii}=\operatorname{trace} (A^2) \cdot \operatorname{trace} (B^2)$$

$^\color{blue}{\star}$ We have applied the Cauchy-Schwarz inequality.
$^\color{red}{\star\star}$ Normally, it is:
$c_{ii} = a_{i1}b_{1i} + a_{i2}b_{2i} +\cdots + a_{in}b_{ni},$
but notice that $b_{k\ell} = b_{\ell k}$.
