Question. Let $a,b,c\ge 0: ab+bc+ca=1.$ Prove that $$\color{black}{\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{c+b}}+\frac{1}{\sqrt{a+c}}\ge 2+\sqrt{\frac{2}{3a+3b+3c-2}}. }$$ I've tried to use Jichen lemma without success.
Indeed, $$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge 2+\frac{2}{3(a+b+c)-2}$$is already wrong when $a=b=c=1/\sqrt{3}.$
Also, I hope the following inequality will help$$\frac{1}{\sqrt{a+b}}+\frac{1}{\sqrt{c+b}}+\frac{1}{\sqrt{a+c}}\ge 2+\frac{\sqrt{2}}{2},$$which gives a proof for OP when $a+b+c\ge 2.$
The rest is proving in remain case $\sqrt{3}\le a+b+c<2.$ I still can not go further.
Hope you help me continue my idea. Is there any better approach to solve the problem?
Thank you for your interest!
If $c=0$, so $ab=1$ and
$$\sum_{cyc}\frac{1}{\sqrt{a+b}}\geq2+\sqrt{\frac{2}{3(a+b+c)-2}}$$ it's $$\frac{1}{\sqrt{a+b}}+\sqrt{a}+\sqrt{b}\geq2+\sqrt{\frac{2}{3(a+b)-2}},$$ which is true because by AM-GM$$\sqrt{a}+\sqrt{b}\geq2$$ and $$\frac{1}{\sqrt{a+b}}\geq\sqrt{\frac{2}{3(a+b)-2}}.$$ Now, let $abc\neq0$, $$f(a,b,c,\lambda)=\sum_{cyc}\frac{1}{\sqrt{a+b}}-2-\sqrt{\frac{2}{3(a+b+c)-2}}+\lambda(ab+ac+bc-1)$$ and $(a,b,c)$ be an inside minimum point of $f$.
Thus, $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0,$$ which gives $$-\tfrac{1}{\sqrt{(a+b)^3}}-\tfrac{1}{\sqrt{(a+c)^3}}+2\lambda(b+c)=-\tfrac{1}{\sqrt{(a+b)^3}}-\tfrac{1}{\sqrt{(b+c)^3}}+2\lambda(a+c)=$$ $$=-\tfrac{1}{\sqrt{(a+c)^3}}-\tfrac{1}{\sqrt{(b+c)^3}}+2\lambda(a+b).$$ Now, let in this point $a\neq b$ and $a\neq c$.
Thus, $$\frac{1}{\sqrt{(b+c)^3}}-\frac{1}{\sqrt{(a+c)^3}}=2\lambda(a-b)$$ or $$\frac{(a+c)^2+(a+c)(b+c)+(b+c)^2}{\sqrt{(a+c)^3(b+c)^3}\left(\sqrt{(a+c)^3}+\sqrt{(b+c)^3}\right)}=2\lambda.$$ By the similar way we obtain: $$\frac{(a+b)^2+(a+b)(b+c)+(b+c)^2}{\sqrt{(a+b)^3(b+c)^3}\left(\sqrt{(a+b)^3}+\sqrt{(b+c)^3}\right)}=2\lambda,$$ which gives $$\tfrac{(a+c)^2+(a+c)(b+c)+(b+c)^2}{\sqrt{(a+c)^3}\left(\sqrt{(a+c)^3}+\sqrt{(b+c)^3}\right)}=\tfrac{(a+b)^2+(a+b)(b+c)+(b+c)^2}{\sqrt{(a+b)^3}\left(\sqrt{(a+b)^3}+\sqrt{(b+c)^3}\right)}$$ or $$(a^2+b^2+ab+3c(a+b+c))(a+b)^3-(a^2+c^2+ac+3b(a+b+c))(a+c)^3+$$ $$+\tfrac{\sqrt{(b+c)^3}\left((a^2+b^2+ab+3c(a+b+c))^2(a+b)^3-(a^2+c^2+ac+3b(a+b+c))^2(a+c)^3\right)}{(a^2+b^2+ab+3c(a+b+c))\sqrt{(a+b)^3}+(a^2+c^2+ac+3b(a+b+c))\sqrt{(a+c)^3}}=0$$ or $$(b-c)\left(\sum_{cyc}(a^4+4a^3b+4a^3c+7a^2b^2+16a^2bc)+\tfrac{\sqrt{(b+c)^3}\left(-a^6-3a^5b-3a^5c-3a^4bc+S\right)}{(a^2+b^2+ab+3c(a+b+c))\sqrt{(a+b)^3}+(a^2+c^2+ac+3b(a+b+c))\sqrt{(a+c)^3}}\right)=0,$$ where $$S=7c^5cb(c^4+b^4)+22b^2c^2(b^2+c^2)31b^3c^3+5a(c^5+b^5)+35abc(b^3+c^3)+$$ $$+77ab^2c^2(b+c)+12a^2(b^4+c^4)+51a^2bc(b^2+c^2)+75a^2b^2c^2+$$ $$+9a^3(b^3+c^3)+21a^3bc(b+c)>0.$$ We'll prove that $$\sum_{cyc}(a^4+4a^3b+4a^3c+7a^2b^2+16a^2bc)+\tfrac{\sqrt{(b+c)^3}\left(-a^6-3a^5b-3a^5c-3a^4bc+S\right)}{(a^2+b^2+ab+3c(a+b+c))\sqrt{(a+b)^3}+(a^2+c^2+ac+3b(a+b+c))\sqrt{(a+c)^3}}>0,$$ for which it's enough to prove that: $$(a+b+c)^4\sqrt{a^3}(2a^2+4b^2+4c^2+4ab+4ac+6bc)\geq\sqrt{(b+c)^3}a^4(a^2+3(ab+ac+bc))$$ or $$4(a+b+c)^8(a^2+2b^2+2c^2+2ab+2ac+3bc)^2\geq(b+c)^3a^5(a^2+3(ab+ac+bc))^2,$$ for which it's enough to prove that: $$4(a+b+c)^{12}\geq9a^5(b+c)^3(a+b)^2(a+c)^2$$ and since by AM-GM $$\prod_{cyc}(a+b)^2\leq\left(\frac{\sum\limits_{cyc}(a+b)}{3}\right)^6=\frac{64(a+b+c)^6}{729}$$ and
$$a(b+c)\leq\frac{(a+b+c)^2}{4},$$ it's enough to prove that: $$81(a+b+c)^4\geq4a^4,$$ which is obvious.
Thus, $b=c$ and it's enough to prove our inequality for equality case of two variables.
Can you end it now?