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sinA + sinB + sinC <= (3 x 3^0.5 )/2
how do we prove it?
A,B,C are angles of a triangle.
thanks for any help.
how do we prove it?
A,B,C are angles of a triangle.
thanks for any help.
sin x is concave, and for concave functions Jensen's inequality is reversed giving:A function is convex (or concace up) if its second derivative is greater than zero. For the full definition of convexity see http://mathworld.wolfram.com/ConvexFunction.html.
Jensens inequality states that the arithmetic mean of a convex function is greater or equal than the function of the arithmetic mean, ie:
[tex]\frac{f(x_1)+f(x_2)+...+f(x_n)}{n} \geq f(\frac{x_1+x_2+...+x_n}{n})[/tex]
So consider the function sinx, and show that sinx is convex. After that, set up jensens inequality and use that the sum of the angles in a triangle is pi.
Asch, sorry, forgot how the original inequality looked like . Thanks!sin x is concave, and for concave functions Jensen's inequality is reversed giving:
[tex]\frac{f(x_1) + f(x_2) + \ldots + f(x_n)}{n} \leq f\left(\frac{x_1+x_2+...+x_n}{n}\right)[/tex]