Suppose, for positive reals $a$, $b$, $c$, that $$ab+bc+ca+abc=4$$ Prove that $$\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\leq3$$
I applied AM-GM on the first equality ie, $a$, $b$, $c$, and $abc$ to get $$ab+bc+ca \geq 3\qquad\text{and}\qquad abc \leq1$$
The exact equation is as follows
$$1=\frac{ab+bc+ca+abc}{4}\ge\sqrt[4]{(abc)^3}\implies 1\ge abc$$
However, I didn't manage to get any further than this after applying AM-GM to several other inequalities.
I'd like a solution for this that utilizes AM-GM only, as I'm very new to inequalities.
Let $a=\frac{2x}{y+z}$ and $b=\frac{2y}{x+z}$, where $x$, $y$ and $z$ be positives.
Thus, the condition gives $$\frac{4xy}{(x+z)(y+z)}+2c\left(\frac{x}{y+z}+\frac{y}{x+z}\right)+\frac{4xyc}{(x+z)(y+z)}=4$$ or $$\frac{2c(x^2+y^2+xz+yz+2xy)}{(x+z)(y+z)}=4-\frac{4xy}{(x+z)(y+z)}$$ or $$\frac{2c(x+y)(x+y+z)}{(x+z)(y+z)}=\frac{4z(x+y+z)}{(x+z)(y+z)}$$ or $$c=\frac{2z}{x+y}$$ and we need to prove that $$2\sum_{cyc}\sqrt{\frac{xy}{(x+z)(y+z)}}\leq3, $$ which is AM-GM: $$2\sum_{cyc}\sqrt{\frac{xy}{(x+z)(y+z)}}\leq\sum_{cyc}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)=$$ $$=\sum_{cyc}\left(\frac{x}{x+z}+\frac{z}{z+x}\right)=3. $$ Done!
There are proofs by trigonometry and $uvw$ but they are not easy.