Understanding Cauchy-Schwarz inequality for matrices

C-S inequality says $$|x_1.y_1+x_2.y_2+\cdots+x_r.y_r|^2\le|x_1^2+x_2^2+\cdots+x_r^2| \cdot |y_1^2+y_2^2+\cdots+y_r^2|$$

Now put $x_i=1$ and $y_i=\lambda_i$.

Let $V$ be a vector space over the the complex numbers and let $\langle \cdot , \cdot \rangle : V \times V \rightarrow \mathbb{C}$ be an inner product map. Note that the map $\left\| \cdot \right\|:V \rightarrow \mathbb{R}^+$ given by $v \mapsto \left\|v\right\| = \langle v,v \rangle^{1/2}$ is a norm map. Then the Cauchy-Schwarz inequality says that for any $u, v \in V$ we have that $\left|\langle u,v \rangle\right| \le \left\|u\right\| \left\|v\right\|$ or $\langle u,v \rangle^2 \le \langle u,u \rangle \langle v,v \rangle$. Let $M_k(\mathbb{C})$ denote the $k \times k$ matrices over the complex numbers. Then the map $M_k(\mathbb{C}) \times M_k(\mathbb{C}) \rightarrow \mathbb{C}$ given by $(A,B) \mapsto Tr(AB^*)$ is an inner product (verify this) and its induced norm is the Frobenius norm. The Cauchy-Schwarz inequality becomes $Tr(AB^*)^2 \le Tr(AA^*) \cdot Tr(BB^*)$. Now take $A$ to be Hermitian and $B$ to be the identity. This gives $\left(Tr(A)\right)^2 \le k Tr(A^2)$, since $Tr(I_k I_k^*) = Tr(I_k) = k$.