Don't Use Calculus! Try Jensen's inequality or AM-GM ;)
Knowing f ( x ) = 2 sin x + sin 2 x , what is the minimum value for f ( x ) ?
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1 solution
You are using the fact that sin x is concave to apply Jensen's Inequality. This is assuming that 0 ≤ x ≤ π , since sin x is not concave (it is convex) between π and 2 π .
Since you are assuming that x and π − 2 x lie in [ − 0 , π ] , all you have proved is that f ( x ) ≤ 2 1 3 3 for 0 ≤ x ≤ 2 1 π . What about other values of x ?
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Sin x has a period of 2pi meaning it can reach at least one extrema in [0, pi], while sin 2x has a period of pi, meaning it can reach at least 3 extrema in [0, pi]. Thus setting 0 =< x =<pi is enough to find one extrema for f(x) (could be Maximum or minimum). Since it is also an odd function, we know the value for the minimum is the negative of the absolute value of the extrema.
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Information about the two functions separately is not enough to be sure you know how many extrema the sum function has, or that you can be sure that one of them occurs in 0 l e x ≤ 2 1 π .
If you are going to handle this, you can observe that f ( π − x = 2\sin x - \sin2x), so that ∣ f ( π − x ) ∣ ≤ f ( x ) for 0 ≤ x ≤ 2 1 π , which means that any extrema of f ( x ) over 2 1 π ≤ x ≤ π are smaller in modulus that the maximum at x = 3 1 π . You can can now use the oddness of f to extend these observations to − π ≤ x ≤ 0 . Since we now know the extrema of f over − π ≤ x ≤ π , we can use the periodicity of f to finish up.
Bottom line - Jensen's Inequality is not enough to solve the whole problem! To that end, a calculus approach is much more straightforward, since we have to do quite a bit of work to make the J.I. argument wholly valid.
∵ f ( x ) = 2 sin x + sin 2 x ⇒ f ( x ) = sin x + sin x + sin ( π − 2 x ) Also ∵ According to the property of sin and cos function that they all starts concaving down before reaching the turning point. ∴ using Jensen's Inequality n f ( x 1 ) + f ( x 2 ) + ⋯ + f ( x n ) ≤ f ( n x 1 + x 2 + ⋯ x n ) sin x + sin x + sin ( π − 2 x ) ≤ 3 ⋅ f ( 3 π ) sin x + sin x + sin ( π − 2 x ) ≤ 2 3 3 ∵ This is an odd function 2 3 3 ≤ sin x + sin x + sin ( π − 2 x )