Inequality without BW and Uvw, just AM GM Problem: Let $a,b,c\ge0: ab+bc+ca=1.$ Prove that: $$(a+b+c)(3-\sqrt{ab}-\sqrt{bc}-\sqrt{ca})+2\sqrt{abc}(\sqrt{a}+\sqrt{b}+\sqrt{c})\ge4$$
I guess equality holds: two of them equal $1$ and one equal to $0$.
My approach: The equality equivalent to
$(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}-a-b-c)+3(a+b+c)\ge5$
I tried to use Pqr method but it is quite complicated. Any one help me to get nice proof? Thanks.

$uvw$ helps!
Let $\sqrt{ab}=z$, $\sqrt{ac}=y$ and $\sqrt{bc}=x$.
Thus, $x^2+y^2+z^2=1$ and we need to prove that:
$$\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}\right)(3-x-y-z)+2(xy+xz+yz)\geq4.$$
Now, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, the condition does not depend on $w^3$, but
$$\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}=\frac{9v^4-6uw^3}{w^3}=\frac{9v^4}{w^3}-6u$$ decreases respect to  $w^3,$ which by $uvw$ says
(about $uvw$ see here:
https://artofproblemsolving.com/community/c6h278791 )  that it's enough to prove the last inequality for equality case of two variables.
Can you end it now?
