<s>[INST] If $ab+bc+ca+abc=4$ then $\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\leq3$, via AM-GM 
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.
 [/INST] 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.
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