Given that $$tr \Bigl( \sum_{i=1}^k (\alpha A_{i} + B_{i})^2 \Bigr) \geqslant 0 \forall \alpha \in \mathbb{R} $$ with $A_{i}$ and $B_{i}$ being positive definite $n\times n$ complex matrices $\forall i=1,...k$
How can I deduce that $\bigg( tr \Bigl( \sum_{i=1}^k A_{i}B_{i}\Bigr) \bigg)^2$ $\leqslant$ $tr \Bigl( \sum_{i=1}^k A_{i}^2 \Bigr)tr \Bigl( \sum_{i=1}^k B_{i}^2 \Bigr)$
I tried substituting -1 instead of $\alpha$ and expanding yet I'm stuck.
Replace $\alpha$ by $-\alpha$ in your first expression. After expanding it, you get $$ \def\tr{\text{tr}} -\alpha\sum \tr A_i^2-\frac1\alpha\sum\tr B_i^2+2\sum\tr (A_iB_i)\leq 0,\\ \sum\tr (A_iB_i)\leq\frac12\left( \alpha\sum \tr A_i^2+\frac1\alpha\sum\tr B_i^2\right) $$ for $\alpha>0$. Then choose $\alpha$ in such a way that $\alpha\sum \tr A_i^2=\frac1\alpha\sum\tr B_i^2$. Apply the inequality $\frac{a^2+b^2}{2}\ge ab$ where $a= \alpha\sum \tr A_i^2$ to get the result.
Alternatively, you can just prove that it is an inner product space. Let $X$ be the set of $k$-tuples of $n\times n$ symmetric matrices. Let $\mathbf A=(A_1,\ldots,A_k)$, and similar for $\mathbf B$. $\mathbf A,\mathbf B\in X$. We consider the function $$ \langle \mathbf A,\mathbf B\rangle:=\tr\left(\sum_{i=1}^k A_iB_i\right). $$ We can verify that $\langle \cdot,\cdot\rangle$ is an inner product. This makes $X$ an inner product space. Then apply the Cauchy-Schwarz inequality, which is true for all inner product spaces.
Of course, $X$ has other different inner products, since $X\subseteq\mathbb R^{kn^2}$. The inner product I described above, however, is one of a few inner products that are "bilinear" for matrices, i.e. for all matrices $M$, $$ \langle M\mathbf A,\mathbf B\rangle=\tr M \cdot\langle \mathbf A,\mathbf B\rangle, \langle \mathbf A,M\mathbf B\rangle=\tr M \cdot\langle \mathbf A,\mathbf B\rangle $$