Can I not use QM-AM inequality to solve this?

Following the hint given in the comments, consider the following diagram:

The inequality's LHS is the sum of lengths of diagonals that have been drawn in the diagram. However, this being a path from corner to opposite corner, it must be at least as long as a straight line, which has length equal to the RHS. This completes the proof.

Here is a purely visual proof based on Michal Adamaszek's comment:

By CS:

$\left(\sum\limits_{cyc}\sqrt{a^2+b^2}\right)^2=2\sum\limits_{cyc}a^2+2\sum\limits_{cyc}\sqrt{a^2+b^2}\sqrt{c^2+a^2}\geq 2\sum\limits_{cyc}a^2+2\sum\limits_{cyc}(ac+ab)=2(a+b+c)^2$