Let $O$ denote the centre of the circle, and let $\alpha, \beta, \gamma$, respectively, be the radian measures of the vertex angles $\angle A, \angle B, \angle C$. Since the circumradius of $ABC$ is $1$, the formula for the circumradius of any triangle tells us that
$$\frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma} = 2,$$
and so the area of $ABC$ is given by
$$(ABC) = \frac{1}{2}ab \sin \gamma = 2 \sin \alpha \sin \beta \sin \gamma.$$

We now show that the function $f(x) = \ln(\sin x)$ is strictly concave on $(0, \pi)$ which enables us to apply Jensen's inequality. Indeed, if $x, y \in (0, \pi)$, then
$$\begin{aligned} \sin x \cdot \sin y &= \frac{1}{2}(\cos(x-y) - \cos(x+y)) \\ &= \sin^2\left(\frac{x+y}{2}\right) - \sin^2\left(\frac{x-y}{2}\right) \le \sin^2\left(\frac{x+y}{2}\right), \end{aligned}$$
with equality iff $x = y$. Taking logarithms this yields
$$\frac{f(x) + f(y)}{2} \le f\left(\frac{x+y}{2}\right)$$
with equality iff $x = y$. This shows that $f$ is strictly concave on $(0, \pi)$ and we can use Jensen's inequality: $f(x) + f(y) + f(z) \le 3f\left(\frac{x+y+z}{3}\right)$, which translates into
$$(ABC) \le 2 \sin^3 \left( \frac{\alpha + \beta + \gamma}{3} \right) = 2 \sin^3 \left( \frac{\pi}{3} \right) = \frac{3\sqrt{3}}{4},$$
with equality iff $\alpha = \beta = \gamma$. That is, $(ABC) \le \frac{3\sqrt{3}}{4}$, with equality iff the triangle is equilateral.
From the formula for the circumradius, already employed above, we obtain $a = 2\sin\alpha$, $b = 2\sin\beta$, and $c = 2\sin\gamma$. Hence the perimeter $p$ of $ABC$ is equal to $2(\sin\alpha + \sin\beta + \sin\gamma)$ and Jensen's inequality for the strictly concave function $\sin(x)$ on $[0, \pi]$ gives
$$p = 2(\sin \alpha + \sin \beta + \sin \gamma) \le 6 \sin \left( \frac{\alpha + \beta + \gamma}{3} \right) = 6 \sin \frac{\pi}{3} = 3\sqrt{3},$$
with equality iff $\alpha = \beta = \gamma = \pi/3$, i.e. equality holds, iff $ABC$ is equilateral.