Prove that$ \frac{a+b}{\sqrt{ab(1-ab)}} + \frac{b+c}{\sqrt{bc(1-bc)}} + \frac{c+a}{\sqrt{ca(1-ca)}} \le \frac{\sqrt{2}}{abc}$ Prove the following inequality where positive reals $a$, $b$, $c$ satisfies $ab+bc+ca=1$.
$$
        \frac{a+b}{\sqrt{ab(1-ab)}} + \frac{b+c}{\sqrt{bc(1-bc)}} + \frac{c+a}{\sqrt{ca(1-ca)}} \le \frac{\sqrt{2}}{abc}
    $$
I got that $ \frac{a+b}{\sqrt{ab(1-ab)}}=\frac{a+b}{\sqrt{ab(bc+ca)}}=\frac{a+b}{\sqrt{abc(a+b)}}.$
So rationalising, we get  $\frac{a+b}{\sqrt{abc(a+b)}}=\frac{(a+b){\sqrt{abc(a+b)}}}{abc(a+b)}=\frac{\sqrt{abc(a+b)}}{{abc}}$
Then enough to show that $\sum_{cyc}\sqrt{abc(a+b)}\le \sqrt2.$
Any help?

let $x=ab,y=bc,z=ca$  and $x+y+z=1$
after your work  we have to show
$$\sum_{cyc} \sqrt{2x(y+z)}\le 2$$ which is true by AM-GM $$\sum_{cyc} \sqrt{2x(y+z)}\le \sum_{cyc}\frac{2x+y+z}{2}=2$$
