sin 2 ( θ 1 ) + sin 2 ( θ 2 ) + ⋯ + sin 2 ( θ 1 0 0 ) [ 1 sin ( θ 1 ) + 2 sin ( θ 2 ) + ⋯ + 1 0 0 sin ( θ 1 0 0 ) ] 2
Let θ 1 , θ 2 , ⋯ , θ 1 0 0 be 100 positive real numbers.
Find the maximum value of the expression above.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Cool!
Can you explain more this case of Cauchy-Schwartz-inequality, please? Thank you!
Just apply Cauchy inequality,for the numerator and the denominator will get cancelled,leaving 1 + 2 + 3 + . . . 1 0 0 = 5 0 5 0 .
Correct.
Problem Loading...
Note Loading...
Set Loading...
This is a oral question if you apply CS inequality appropriately.. Let S denote the given expression. Then by Cauchy-Schwarz-Inequality S ≥ sin 2 ( θ 1 ) + sin 2 ( θ 2 ) + ⋯ + sin 2 ( θ 1 0 0 ) ( 1 + ( 2 ) 2 + ⋯ ( 1 0 0 ) 2 ) [ sin 2 ( θ 1 ) + sin 2 ( θ 2 ) + ⋯ + sin 2 ( θ 1 0 0 ) = 1 + 2 + 3 + ⋯ 1 0 0 = 2 1 0 0 ( 1 0 1 ) = 5 0 5 0
Equality occurs when: 1 sin θ 1 = 2 sin θ 2 = ⋯ = 1 0 0 sin θ 1 0 0