Let $D$ be the point on $AB$ such that $\angle ACD=\frac{\pi}{2}$. So $CD=CA$ and $AD=\sqrt{2}CA$. By triangle inequality on $BCD$, we have $BD+CD>BC$, so $AB=AD+BD=\sqrt{2}CA+BD>\sqrt{2}CA+BC-CD>BC+(\sqrt{2}-1)CA$

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