I need to find triangles for which is sinx*siny*sinz maximum when x,y,z are angles. Are those ones with all three angles 60°?
True
Proof:
$\displaystyle \sqrt[3]{\sin x *\sin y* \sin z}\leq \frac{\sin x + \sin y + \sin z}{3}\leq \sin \left(\frac{x+y+z}{3}\right)$
the first inequality: the geometric average is less than the arithmetic average
the second one you can prove using the fact that the graph of sin is concave down between 0 and pi
in our situation x + y + z = pi, so
$\displaystyle \sin x *\sin y* \sin z\leq \left(\left.\sqrt{3}\right/2\right)^3$
Hi,
Idea's proof is very clever, but it is "advanced". You need to know the inequality about means (standard, but no obvious proof) and some basic knowledge about convex sets. Assuming you know elementary calculus, you can solve the problem easily with Lagrange multipliers:
Maximize $F(x,y,z)=\sin(x)\cdot \sin(y)\cdot \sin(z)$ subject to the constraint $x+y+z=\pi$ where each of $x,y,z$ are in $[0,\pi]$.
Furthermore this method finds the solution for you.