Show that $\text{trace}(A)\leq \sqrt{n}\sqrt{\text{trace}(AA^t)}$ where $A$ is in $M_{n}(\mathbb{R}).$
We know that $\text{trace}(A) =\sum_{i}\lambda_i$ where $\lambda_i$ are eigenvalues of $A.$ Then I was wondering if we could show that the eigenvalues of $AA^t$ are $\lambda_i^2$ then we could apply the AM-GM inequality to complete the proof. So we know that $A$ and $A^t$ have the same eigenvalues, but can we use this conclude that $AA^t$ have $\lambda_i^2$ as eigenvalues?
No, the eigenvalues of $AA^t$ are not necessarily the squared eigenvalues of $A$. E.g. when $A=\pmatrix{0&1\\ 0&0}$, all eigenvalues of $A$ are zero, but the eigenvalues of $AA^t=\pmatrix{1&0\\ 0&0}$ are $1$ and $0$.
At any rate, the inequality in question is just Cauchy-Schwarz inequality. Note that $\langle X,Y\rangle:=\operatorname{tr}(XY^T)$ is an inner product on $M_n(\mathbb R)$ and the inequality can be rewritten as $\langle A,I\rangle^2 \le \langle I,I\rangle\langle A,A\rangle$.