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:).

Here is a possible approach without using Hölder's inequality. Assume without loss of generality that $a\ge b \ge c$. Then
$$\left({a\over {\sqrt {b+c}}}+{b\over {\sqrt {c+a}}}+{c\over {\sqrt {a+b}}}\right)^2\ge \frac{1}{2a}\left(a+b+c\right)^2$$
Now
$$ab+bc+ca=A=\frac{A+A}{2}=\frac{1}{2}(a(b+c)+b(c+a)+c(a+b))\ge c(a+b+c)$$
Since $a,b,c\in \mathbb{R}_+$ this implies
$$\left({a\over {\sqrt {b+c}}}+{b\over {\sqrt {c+a}}}+{c\over {\sqrt {a+b}}}\right)^2(ab+bc+ca) \ge \frac{c}{2a}(a+b+c)^3$$
However $0<\frac{c}{a}\le 1$ by assumption, so the desired inequality follows.
