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: Consider the polynomial $$
\begin{align}p(x) &= (x-a)(x-b)(x-c)\\
&= x^3 - Ax^2 + Bx - C\end{align}\quad\text{ where }\quad
\begin{cases} A = a + b + c \\B = ab+bc+ca \\C = abc\end{cases}$$ 
Notice 
$$(a+b)(b+c)(c+a) = (A-c)(A-a)(A-b) = p(A) = BA - C$$
We have
$$(a+b)(b+c)(c+a) = 1 \iff BA - C = 1 \iff B = \frac{1+C}{A}$$
Apply AM $\ge$ GM to the three pairs $(a,b), (b,c), (c,a)$ and combine the result, we obtain
$$8C = (2\sqrt{ab})(2\sqrt{bc})(2\sqrt{ca}) \le (a+b)(b+c)(c+a) = 1$$
Apply AM $\ge$ GM again to the 3-tuple $(a+b,b+c,c+a)$, we find
$$\frac23 A = \frac13((a+b) + (b+c) + (c+a)) \ge \sqrt[3]{(a+b)(b+c)(c+a)} = 1$$
These mean $C \le \frac18$ and $A \ge \frac32$. Combine them, we can deduce
$$ab+bc+ca = B = \frac{1+C}{A} \le \frac{1 + \frac18}{\frac32} = \frac34$$
A: $$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)$$ 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:
$$(a+b+c)^2=\sum_{cyc}(a^2+2ab)=\frac{1}{2}\sum_{cyc}(a^2+b^2+4ab)\geq\frac{1}{2}\sum_{cyc}(2ab+4ab)=3\sum_{cyc}ab.$$
