SinA + sinB + sinC <= (3 x 3^0.5 )/2

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  • #1
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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.
 

Answers and Replies

  • #3
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sorry kurret i am still unable to prove it
 
  • #4
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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]
For concave functions the inequality is reversed. So consider the function sinx, and show that sinx is concave. After that, set up jensens inequality and use that the sum of the angles in a triangle is pi.
 
Last edited:
  • #5
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Alternative solution:
Rewrite C as pi-A-B, then Sin(C)=Sin(A+B). Now assume that B is any value between 0 and pi, and you can make the LHS a function of A, and find its minimum value. Maybe easier, but I think jensens inequality is really powerful and its really a good idea to learn to master it :)
 
  • #6
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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.
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]
 
  • #7
143
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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]
Asch, sorry, forgot how the original inequality looked like :frown:. Thanks!
 

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