Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Prove that$$\color{black}{\sqrt{a+b}+\sqrt{c+b}+\sqrt{a+c}\le a+b+c+\frac{1}{2}abc+\sqrt{2}.}$$ Since $abc\ge 0$ and equality holds at $a=b=1;c=0,$ I've tried to prove stronger$$\sqrt{a+b}+\sqrt{c+b}+\sqrt{a+c}\le a+b+c+\sqrt{2}.$$ But it's wrong when $a=b=c=1/\sqrt{3}.$
Also, here is an idea about Mixing variables technique.
WLOG, assume that $a\ge b\ge c\ge 0.$ Chose $t>0$ such that $t^2+2tc=1=ab+bc+ca\implies (t+c)^2=c^2+1.$\ We get $a\ge t\ge b\ge c$ and $1\ge t^2\implies 0< t\le 1.$\ Now, we need to show $$f(a,b,c)\le f(t,t,c).$$ I was stuck here because it is hard for me to use MV in radical inequality.
Hope to see some ideas. All contributions are welcome.
By C-S twice we obtain: $$\sum_{cyc}\sqrt{a+b}\leq\sqrt{\sum_{cyc}\frac{a+b}{(\sqrt2-1)c+a+b}\sum_{cyc}((\sqrt2-1)c+a+b)}=$$ $$=\sqrt{(\sqrt2+1)(a+b+c)\left(3-(\sqrt2-1)\sum_{cyc}\frac{c}{(\sqrt2-1)c+a+b}\right)}\leq$$ $$\leq\sqrt{(\sqrt2+1)(a+b+c)\left(3-\frac{(\sqrt2-1)(a+b+c)^2}{(\sqrt2-1)(a^2+b^2+c^2)+2}\right)}$$ and it's enough to prove that $$\sqrt{(\sqrt2+1)(a+b+c)\left(3-\tfrac{(\sqrt2-1)(a+b+c)^2}{(\sqrt2-1)(a^2+b^2+c^2)+2}\right)}\leq a+b+c+\frac{1}{2}abc+\sqrt2.$$ The rest is $uvw$ because the left hand side does not depend on $w^3=abc$.
Can you end it now?