We apply AM$\geq$GM
$\frac{a+b}{2}\geq \sqrt{ab}$ ...1
$\frac{b+c}{2}\geq \sqrt{bc}$ ...2
$\frac{a+c}{2}\geq \sqrt{ac}$ ...3
Multiply 1,2,3
$\frac{(a+b)(b+c)(c+a)}{8}\geq abc$
$(a+b)(b+c)(c+a)\geq 8abc$
Now we have to prove that abc$\leq 1$ for equation to hold
again we apply AM$\geq$GM
$\frac{a+b+c}{3}\geq abc^{\frac{1}{3}}$ ...4
we have abc(a+b+c)=3
$(a+b+c)=\frac{3}{abc}$
substituting above result in equation 4 we get
$\frac{3}{\frac{abc}{3}}\geq abc^{\frac{1}{3}}$
$1\geq abc^{\frac{4}{3}}$
hence if $1\geq abc^{\frac{4}{3}}$ then $abc \leq 1$
