By Heron’s formula for the area of the triangle,

$\displaystyle \begin{array}{rcl}S &=& \sqrt{2abc(2abc-a)(2abc-b)(2abc-c)}\\\\

{} &=& abc\sqrt{2(2ab-1)(2bc-1)(2ca-1)}\\\\

{} &=& abc\sqrt{2\left[8(abc)^2-4abc(a+b+c)+2(ab+bc+ca)-1\right]}\\\\

{} &=& abc\sqrt{2\left[4(abc)^2+2(ab+bc+ca)-1\right]}\end{array}$

By AM–GM, $\displaystyle ab+bc+ca\geq3(abc)^{\frac23}.$

$\displaystyle \therefore\ S^2\geqslant2(abc)^2\left[4(abc)^2+6(abc)^{\frac23}-1\right]=2x^3(4x^3+6x-1)$ where $\displaystyle x=(abc)^{\frac23}.$

But, again by AM–GM, $\displaystyle abc=a+b+c\geqslant3(abc)^{\frac13}$ $\displaystyle \implies$ $\displaystyle (abc)^2\geqslant27$ $\displaystyle \implies$ $\displaystyle x\geqslant3.$

Now, $\displaystyle S>1$ $\displaystyle \iff$ $\displaystyle S^2>1$ $\displaystyle \iff$ $\displaystyle 8x^6+4x^4-12x^3-1>0.$ Thus if you can show that $\displaystyle x\geqslant3\ \Rightarrow\ 8x^6+4x^4-12x^3-1>0,$ you are done. I think this should be straightforward.