How do I find the lengths of the sides of a triangle that is inscribed in the unit circle that has the greatest perimeter? I know how to find the sides of the triangle when seeking to maximize the area of ​​the triangle, since the following relationship exists $Area= \frac{abc}{4R}$ where $a, b,c$ are the sides and $R$ the radius.
But I don't know what to occupy to maximize the perimeter.
 A: Recall that the Law of Sines tells you that
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R,$$
where $R$ is the circumradius.  Then the perimeter is
$$2R(\sin A + \sin B + \sin C).$$
Note that
$$\sin A + \sin B = 2\cos(\tfrac{A - B}{2})\sin(\tfrac{A + B}{2}) \le 2\sin(\tfrac{A + B}{2})$$
with equality only when $A = B$.  Thus, the perimeter is maximized when $A = B$.  Using similar logic, you can show that $B = C$ as well when the perimeter is maximized.  So, the perimeter is maximized when $A = B = C = 60^\circ$ with a value of
$$2R(\sin 60^\circ + \sin 60^\circ + \sin 60^\circ) = 3\sqrt3 R.$$
A: Let $O$ be the circumcentre, and let $\alpha,\beta,\gamma$ be the angles at $O$ opposite the sides of the triangle of lengths $a$, $b$ and $c$ respectively. By the cosine rule, we have
$$
a+b+c=\sqrt{2}R\left(\sqrt{1-\cos \alpha}+\sqrt{1-\cos \beta}+\sqrt{1-\cos \gamma}\right).
$$
Now $\frac{d^2}{dx^2}\left(\sqrt{1-\cos x}\right)=\frac{-1}{4}\sqrt{1-\cos x}\leq 0$, so $\sqrt{1-\cos x}$ is concave. Thus by Jensen's inequality,
$$
\sqrt{1-\cos \alpha}+\sqrt{1-\cos \beta}+\sqrt{1-\cos \gamma}\leq 3\cdot \sqrt{1-\cos \frac{\alpha+\beta+\gamma }{3}}=3\sqrt{1-\cos\frac{2\pi }{3}}
$$
with equality iff $\alpha=\beta=\gamma=\frac{2\pi }{3}\Longleftrightarrow a=b=c$.
