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

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