<s>[INST] if $AB+BC+AC\le 2+\sqrt{3}$ find the maxmum of the value $P=AB\cdot BC\cdot AC$ For $\Delta ABC$  if the circumradius $R=1$,and such
$$AB+BC+AC\le 2+\sqrt{3}$$
find the maxmum of the value
$$P=AB\cdot BC\cdot AC$$
if let $a=BC,b=AC,c=AB$,then we have
$$a+b+c\le 2+\sqrt{3}$$
and if use sine theorem we have
$$a=2R\sin{A}=2\sin{A},b=2\sin{B},c=2\sin{C}$$
so
$$\sin{A}+\sin{B}+\sin{C}\le\dfrac{2+\sqrt{3}}{2}$$
so we want to find the maximum of the value
$$P=AB\cdot BC\cdot AC=abc=8\sin{A}\sin{B}\sin{C}$$
if we use AM-GM
$$\sin{A}\sin{B}\sin{C}\le\left(\dfrac{\sin{A}+\sin{B}+\sin{C}}{3}\right)^3\le\left(\dfrac{2+\sqrt{3}}{2}\right)^3$$
the $=$ iff $A=B=C$ but this is clear wrong.so How to solve this problem? Thanks
 [/INST] We need to find the maximum of $P = 8\sin A \sin B \sin C$ under the conditions
$A, B, C \ge 0$, $A + B + C = \pi$, and $2\sin A + 2\sin B + 2\sin C \le 2 + \sqrt{3}$.
From the conditions, we have
$(2\sin A + 2\sin B)^2 \le (2 + \sqrt{3} - \sin C)^2$.
We have
\begin{align*}
 (2 \sin A + 2\sin B)^2 &= 16 \sin^2 \frac{A + B}{2} \cos^2 \frac{B - A}{2}\\
 &= 4(\cos C + 1)(\cos(B - A)  + 1).
\end{align*}
Then, we have
$$4(\cos C + 1)(\cos(B - A)  + 1) \le (2 + \sqrt{3} - 2\sin C)^2$$
which results in
$$\cos(B - A) \le \frac{(2 + \sqrt{3} - 2\sin C)^2}{4(\cos C + 1)} - 1.$$
On the other hand, we have
\begin{align*}
 P &= 8\sin A \sin B \sin C\\
 &= 4[\cos (B - A) - \cos(B + A)]\sin C\\
 &= 4[\cos (B - A) + \cos C]\sin C.
\end{align*}
We split into two cases:
(1) $C \ge 2\pi/3$:
We have
\begin{align*}
 P &\le 4(1 + \cos C)\sin C \\
 &= 4\left(1 + \frac{1 - u^2}{1 + u^2}\right)\frac{2u}{1 + u^2}\\
 &= \frac{16u}{(1 + u^2)^2}\\
 &\le \sqrt{3}
\end{align*}
where $u = \tan \frac{C}{2} \ge \sqrt{3}$
(with equality if $A = B = \pi/6, \ C = 2\pi/3$).
(2) $C < 2\pi/3$:
We have
\begin{align*}
 P &\le 4\left[\frac{(2 + \sqrt{3} - 2\sin C)^2}{4(\cos C + 1)} - 1 + \cos C \right]\sin C \\ 
 &= -(8\sqrt{3} + 16)\frac{u^2}{1 + u^2} + (4\sqrt{3} + 7) u \\
 &= \sqrt {3}-{\frac { \left( 4\,\sqrt {3}+7 \right)  \left( \sqrt {3} - u
   \right)  \left( u-2+\sqrt {3} \right) ^{2}}{{u}^{2}+1}}\\
 &\le \sqrt{3}
\end{align*}
where $u = \tan \frac{C}{2} \in (0, \sqrt{3})$.
So, the maximum of $P$ is $\sqrt{3}$ at $A = B = \pi/6, \ C = 2\pi/3$.
</s>