How to show that $ab+bc+ca\le \frac34$ Let $a,b$ and $c$ be positive real numbers such that $(a+b)(b+c)(c+a) = 1$ , hen show that $$ab+bc+ca\le \frac34$$
I believe I need to use AM-GM inequality and use the fact $(a+b)(b+c)(c+a) = 1$ Using AM-GM in $a+b,b+c$ & $c+a$, I get $a+b+c\ge \frac32$. Any hint will be thankful. 

$$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$$ it's
$$\sum_{cyc}(a^2b+a^2c-2abc)\geq0,$$ which is true by AM-GM:
$$\sum_{cyc}(a^2b+a^2c-2abc)=\sum_{cyc}(a^2c+b^2c-2abc)\geq\sum_{cyc}(2\sqrt{a^2c\cdot b^2c}-2abc)=0.$$
Thus, it's enough to prove that
$$(a+b+c)^2\geq3(ab+ac+bc),$$ which is true by AM-GM again:
$$(a+b+c)^2=\sum_{cyc}(a^2+2ab)=\frac{1}{2}\sum_{cyc}(a^2+b^2+4ab)\geq\frac{1}{2}\sum_{cyc}(2ab+4ab)=3\sum_{cyc}ab.$$
