An Olympiad Inequality Problem

Since $abc=1$ , we have

$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-1$

Hence, it suffices to prove that

$(a+b+c)(ab+bc+ca)-1 \geq N(a+b+c-1)$

or

$ab+bc+ca+\frac{3}{a+b+c} \geq N.$

We shall use the inequality

$(x+y+z)^2 \geq 3(xy+yz+zx),$

which after expansion and cancelling common terms yields

$\frac{1}{2}\left[(x-y)^2+(y-z)^2+(z-x)^2\right] \geq 0.$

It is easy to see that

$(ab+bc+ca)^2 \geq 3(ab \times bc + bc \times ca + ca \times ab) = 3(a+b+c).$

By AM-GM,

$ab+bc+ca+\frac{3}{a+b+c}$

$=3\left(\frac{ab+bc+ca}{3}\right)+\frac{3}{a+b+c}$

$\geq 4 \sqrt[4]{\frac{3(ab+bc+ca)^3}{3^3(a+b+c)}}$

$\geq 4 \sqrt[4]{\frac{3(3\sqrt[3]{a^2b^2c^2})[3(a+b+c)]}{3^3(a+b+c)}}=4,$

and we are done. $_\square$

Isn't it easier to just assume a=b=c=1 then you can solve with ease