Proof:

With S - area, p - semiperimeter, r - inradius:

Heron's formula: S*S = p(p - a)(p - b)(p- c).

p(p - a)(p - b)(p- c) <= p{<(p-a) + (p-b) + (p-c)>/3}^3

S^2 <= p(p/3)^3

S <= p^2 / 3sqrt(3)

r 3sqrt(3) <= p.

It fallows that the equlateral traingle has the smallest perimiter and, from S = pr, the smallest area.

Question: a similar statement is probably true for n-gons, with n > 3, so that a regular n-gon has the smallest area among all the n-gons circumscribed around a given circle.

tan α + tan β + ... ≥ n·tan(180°/n).

But, since tan is concave on (0, π/2),

(tan α + tan β + ...)/n ≥ tan{(α + β + ...)/n}

and the assertion follows right away.