Let $\mathcal{A}$ be a $C^{\ast}$-algebra and $a_1, a_2, b_1$ and $b_2$ nonzero elements of $\mathcal{A}$. Then it is not difficult to see that
$\vert \vert a_1b_1+a_2b_2 \vert \vert^2 \leq (\vert \vert a_1 \vert \vert^2 +\vert \vert a_2 \vert \vert^2 ) (\vert \vert b_1 \vert \vert^2 +\vert \vert b_2 \vert \vert^2) $
Does the above inequality becomes an equality in case $a_1, a_2, b_1$ and $b_2$ are linearly dependent?
No. Pick $x$ to be a nonzero element whose square is $0$. Say, the matrix $\begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}$. Let $a_1 = a_2 = b_1 = b_2 = x$. Then the LHS is $0$ but the RHS is $4\|x\|^4 \neq 0$.