Olympiad inequality problem with $a+b+c+abc=4$ If $a,b,c \in \mathbb R_{> 0}$ and $a+b+c+abc=4$, prove that
$$({a\over {\sqrt {b+c}}}+{b\over {\sqrt {c+a}}}+{c\over {\sqrt {a+b}}})^2(ab+bc+ca) \ge {\frac 12}(4-abc)^3$$
This can be solved by AM-GM-HM or the Cauchy-Schwarz inequality. I'd tried for some hours but couldn't solve it. Can anyone help me? Thanks in advance:).

This inequality can be written like $$(\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}})^2(a(b+c)+b(c+a)+c(a+b))\geq(a+b+c)^3$$ and this is true with Hölder's inequality
