Show that the following inequality holds between the perimeter $p$ and the area of the triangle $a$.
$$p^2 \ge 12\sqrt3\ a$$
Show that the following inequality holds between the perimeter $p$ and the area of the triangle $a$.
$$p^2 \ge 12\sqrt3\ a$$
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Using Heron's formula, we can write the desired inequality as: $$ (2s)^2 \ge 12 \sqrt3 \sqrt{s(s-a)(s-b)(s-c)} $$
which on squaring, simplifies to $$ s^3 \ge 27 (s-a)(s-b)(s-c) $$
Can you recognise the AM-GM now?