Q: Minimum of $abc$ when $a^3+b^3+c^3\geq a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}.$ 
Let $a,b,c$ be positive real numbers with $abc=k$ such that the inequality $$a^3+b^3+c^3\geq a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}$$
holds for all $a,b,c$. Find the minimum value of $k$.

I found that $abc=2$ works. Here is my proof for $abc=2$:
By Cauchy-Schwarz and Muirhead inequality
$$\begin{align} a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b} &\leq \sqrt{2(a^2+b^2+c^2)(a+b+c)}\\ &= \sqrt{abc(a^2+b^2+c^2)(a+b+c)}\\ &= \sqrt{\sum_{cyc}a^4bc+\sum_{cyc}(a^3b^2c+a^2b^3c)}\\ &\leq \sqrt{\sum_{cyc}a^6+2\sum_{cyc}a^3b^3}\\ &= a^3+b^3+c^3
\end{align}$$
as desired. Equality holds when $a=b=c=\sqrt[3]2$.
The inequality seems quite tight also. But I am not sure if it is the minimum of $abc$. And if it is, how do I prove that?

A: Let $abc=k$.
Thus, after homogenization we need to prove that
$$\sum_{cyc}\left(a^3-a\sqrt{\frac{abc(b+c)}{k}}\right)\geq0,$$ which for $a=b=c$ gives $k\geq2$ and it's enough to prove that:
$$\sum_{cyc}\left(a^3-a\sqrt{\frac{abc(b+c)}{2}}\right)\geq0,$$ which you proved already by C-S.
Also, it's true by AM-GM and Muirhead:
$$\sum_{cyc}\left(a^3-a\sqrt{\frac{abc(b+c)}{2}}\right)=\sum_{cyc}\left(a^3-a\sqrt{\frac{2bc(ab+ac)}{4}}\right)\geq$$
$$\geq\sum_{cyc}\left(a^3-\frac{a(2bc+ab+ac)}{4}\right)=\frac{1}{4}\sum_{cyc}(4a^3-a^2b-a^2c-2abc)\geq0.$$
