Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Prove that$$\color{black}{\sqrt{a+b-2ab}+\sqrt{c+b-2bc}+\sqrt{a+c-2ca}}\ge 2\sqrt{a+b+c-1}.$$ Note that by AM-GM inequality $a+b\ge 2\sqrt{ab}\ge 2ab$ since $0\le ab\le 1.$
Thus, I tried to square both side which gives $$2(a+b+c)-2+2\sum_{cyc}\sqrt{(a+b-2ab)(b+c-2bc)}\ge 4(a+b+c-1)$$ Or $$\sum_{cyc}\sqrt{(a+b-2ab)(b+c-2bc)}\ge a+b+c-1.$$ Also, we have $\sqrt{(a+b-2ab)(b+c-2bc)}\ge \sqrt{(a+b-2\sqrt{ab})(b+c-2\sqrt{bc})}=|(\sqrt{a}-\sqrt{b})(\sqrt{b}-\sqrt{c})|.$
It remains to prove $$\sum_{cyc}(\sqrt{a}-\sqrt{b})(\sqrt{b}-\sqrt{c})|+1\ge a+b+c.$$
However, I don't know how to go further. Hope you can give me some ideas. Thank you!
We need to prove that: $$\sum_{cyc}\left(2a-2ab+2\sqrt{(a+b-2ab)(a+c-2ac)}\right)\geq4(a+b+c-1)$$ or$$\sum_{cyc}\sqrt{(a+b-2ab)(a+c-2ac)}\geq a+b+c-1$$ or $$\sum_{cyc}(a+b-2ab)(a+c-2ac)+2\prod_{cyc}\sqrt{a+b-2ab}\sum_{cyc}\sqrt{a+b-2ab)}\geq(a+b+c-1)^2$$ and it's enough to prove that: $$\sum_{cyc}(a+b-2ab)(a+c-2ac)\geq(a+b+c-1)^2$$ or $$\sum_{cyc}(a^2+3ab-2a^2b-2a^2c-4abc+4a^2bc)\geq a^2+b^2+c^2-2(a+b+c)+3$$ or $$a+b+c+2abc(a+b+c)\geq\sum_{cyc}(a^2b+a^2c+abc)+3abc $$ or $$a+b+c+2abc(a+b+c)\geq(a+b+c)(ab+ac+bc)+3abc $$ or $$2(a+b+c)\geq3$$ or $$2(a+b+c)\geq3\sqrt{ab+ac+bc},$$ which is obvious.