Let $A=\left(\begin{array}{l}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b\\ c\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}d\end{array}\right)$, where $a,b,c,d\in \mathbf{R}$ Then

$\begin{array}{l}\det \left(A\right)=ad-bc\le |ad-bc|\\ \le |a\parallel d|+|b\parallel c|\end{array}$

Now, 

$\begin{array}{l}\left(a^2+b^2\right)\left(c^2+d^2\right)-\left(\right|a\parallel d|+|b\parallel c|{)}^2\\ =a^2{c}^2+b^2{c}^2+a^2{d}^2+b^2{d}^2\\ -\left(a^2{d}^2+b^2{c}^2+2|a\parallel d\parallel b\parallel c|\right)\\ =\left(\right|ac|-|bd|{)}^2\ge 0\end{array}$

$\Rightarrow |a\parallel d|+\left|b\right|\left|c\right|\le \sqrt{a^2+b^2}\sqrt{c^2+d^2}$