If $ab+bc+ca+abc=4$, then $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c$ Let $a,b,c$ be positive reals such that $ab+bc+ca+abc=4$.
Then prove
$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c $
So high guys im a high schooler trying to solve this inequality. I did a few things such as try to set a> b>c and then try to generalize it but i couldn't make much progress at all. I would appreciate a clear solution to help me understand the problem in depth.
I did manage to generalize that 
$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq a+b+c $
By using the AM-GM inequality on each of a b and c however havent made much progress in proving that the LHS is less than 3 and the RHS is greater than 3.

AM-GM on $ab$, $bc$, $ca$ and $abc$ gives:
$$1=\frac{ab+bc+ca+abc}{4}\ge\sqrt[4]{(abc)^3}\implies 1\ge abc$$
So, still using the first equation:
$$ab+bc+ca\ge 3$$
Now, by AM-GM
$$a^2+b^2\ge 2ab$$
$$b^2+c^2\ge 2bc$$
$$c^2+a^2\ge 2ac$$
Adding them all up and dividing by $2$ gives:
$$a^2+b^2+c^2\ge ab+bc+ca$$
So now:
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\ge 3(ab+bc+ca)\ge 9$$
From which it follows that:
$$a+b+c\ge 3$$
I'll work on the other inequality when I have time
