Prove that if $a+b+c=1$ then $\sum\limits_{cyc}\frac{1}{\sqrt{a^2+b^2}}\le\frac{9\sqrt{2}}{2}$

Question

Let $a,b,c>0$,and such $a+b+c=1$,prove or disprove $$\dfrac{1}{\sqrt{a^2+b^2}}+\dfrac{1}{\sqrt{b^2+c^2}}+\dfrac{1}{\sqrt{c^2+a^2}}\le\dfrac{9\sqrt{2}}{2}\tag{1}$$

My try:since $$\sqrt{a^2+b^2}\ge\dfrac{\sqrt{2}}{2}(a+b)$$ it suffices to prove that $$\sum_{cyc}\dfrac{1}{a+b}\le\dfrac{9}{2}$$ But since Cauchy-Schwarz inequality we have $$2\sum_{cyc}(a+b)\sum_{cyc}\dfrac{1}{a+b}\ge 9$$ or $$\sum_{cyc}\dfrac{1}{a+b}\ge \dfrac{9}{2}$$ so this try can't works.so How to prove $(1)$

Answer 1

It's wrong. Try $a=b\rightarrow0^+$

Answer 2

It is false. Try $a=\frac{9}{10},\, b=c=\frac{1}{20}$.