It suffices to prove that $\sum_{cyc}\frac{1}{\sqrt{b+c}} \ge \frac{3}{\sqrt{2}}$ because of Chebyshev. Applying AM-HM gives $\sum_{cyc}{\frac{1}{\sqrt{b+c}}} \ge \frac{9}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{a+c}} \ge \frac{3}{\sqrt{2}}$ $3\sqrt{2} \ge \sqrt{a+b} + \sqrt{b+c} + \sqrt{a+c}$ but idk if this is true or helps in any way EDIT: last result isn't true.

parmenides51 wrote: Let $a,b,c>0$ and $a+b+c+abc=4$. Prove that $$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b} }\ge \frac{1}{\sqrt{2}}(a+b+c).$$ (Zhuge Liang) It's the known Cesar Lupu's inequality.

parmenides51 wrote: Let $a,b,c>0$ and $a+b+c+abc=4$. Prove that $$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b} }\ge \frac{1}{\sqrt{2}}(a+b+c).$$ (Zhuge Liang) We have $$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b} } \ge \frac{(a + b + c)^{3/2}}{\sqrt{a(b+c) + b(c+a) + c(a+b)}}.$$

parmenides51 wrote: Let $a,b,c>0$ and $a+b+c+abc=4$. Prove that $$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b} }\ge \frac{1}{\sqrt{2}}(a+b+c).$$ (Zhuge Liang) We have $$a+b+c \geq ab+bc+ca.$$So $$\sum\limits_{cyc} {a\sqrt {\frac{2}{{b + c}}} = \sum\limits_{cyc} {\frac{{2a}}{{\sqrt {2\left( {b + c} \right)} }}} } \ge \sum\limits_{cyc} {\frac{{4a}}{{b + c + 2}}} = \sum\limits_{cyc} {\frac{{4{a^2}}}{{a(b + c + 2)}}} \ge \frac{{2{{\left( {a + b + c} \right)}^2}}}{{ab + bc + ca + a + b + c}} \ge a + b + c.$$

parmenides51 wrote: Let $a,b,c>0$ and $a+b+c+abc=4$. Prove that $$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b} }\ge \frac{1}{\sqrt{2}}(a+b+c).$$ (Zhuge Liang) https://artofproblemsolving.com/community/c6h127956p725874