$\mathbf{The \ Problem \ is}:$ Let $A$ and $B$ be two $2\times 2$ real matrices in $(\operatorname{M}_2(\mathbb{R}),\|\cdot\|)$ where $\|\cdot\|$ is the Frobenius norm in $\operatorname{M}_2(\mathbb{R}).$ Show that $|\operatorname{det}(A)-\operatorname{det}(B)|\leq \alpha (\|A\|+\|B\|)\|A-B\|$ for an $\alpha> 0.$
$\mathbf{My \ approach}:$ I was thinking to explicitely compute $|\operatorname{det}(A)-\operatorname{det}(B)|$ for $A:= (a_{ij})_{2\times 2}$ and $B:= (b_{ij})_{2\times 2}$ and then make some manipulations so that we can indeed reach expressions similar to the right-hand side. But I couldn't exactly proceed like that as the computations are getting much bigger so I tried in a different way : $|\operatorname{det}(A)-\operatorname{det}(B)|\leq |\operatorname{det}(A)|+|\operatorname{det}(B)|\leq \big(\sqrt{a_{11}^2+a_{21}^2}\sqrt{a_{12}^2+a_{22}^2}\big)+\big(\sqrt{b_{11}^2+b_{21}^2}\sqrt{b_{12}^2+b_{22}^2}\big)\leq \frac{\|A\|^2+\|B\|^2}{2}$ using that GM $\leq$ AM. But I need to have $\|A\|+\|B\|$ on the right. The other way I was thinking was to use the Cauchy-Schwarz in-equality repeatedly. Do we need to use the fact that the Frobenius norm is invariant under multiplication with orthonormal columns from the left and invariant under multiplication with orthonormal rows from the right ? I think it's quite easy but still I need a small hint. Thanks in advance for some help!