Maybe a little exaggerated?

Algebra Level 4

[ 1 sin ( θ 1 ) + 2 sin ( θ 2 ) + + 100 sin ( θ 100 ) ] 2 sin 2 ( θ 1 ) + sin 2 ( θ 2 ) + + sin 2 ( θ 100 ) \dfrac{[\sqrt{1}\sin(\theta_{1}) + \sqrt{2}\sin(\theta_{2}) + \cdots + \sqrt{100}\sin(\theta_{100})]^2}{\sin^{2}(\theta_{1}) + \sin^{2}(\theta_{2}) + \cdots + \sin^{2}(\theta_{100})}

Let θ 1 , θ 2 , , θ 100 \theta_{1} , \theta_{2}, \cdots,\theta_{100} be 100 positive real numbers.

Find the maximum value of the expression above.


The answer is 5050.0.

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2 solutions

Rishabh Jain
Mar 26, 2016

This is a oral question if you apply CS inequality appropriately.. Let S \mathfrak S denote the given expression. Then by Cauchy-Schwarz-Inequality S ( 1 + ( 2 ) 2 + ( 100 ) 2 ) [ sin 2 ( θ 1 ) + sin 2 ( θ 2 ) + + sin 2 ( θ 100 ) sin 2 ( θ 1 ) + sin 2 ( θ 2 ) + + sin 2 ( θ 100 ) \mathfrak S\geq \dfrac{(1+(\sqrt 2)^2+\cdots (\sqrt{100})^2)\cancel{\color{#D61F06}{[\sin^{2}(\theta_{1}) + \sin^{2}(\theta_{2}) + \cdots + \sin^{2}(\theta_{100})}}}{\cancel{\color{#D61F06}{\sin^{2}(\theta_{1}) + \sin^{2}(\theta_{2}) + \cdots + \sin^{2}(\theta_{100})}}} = 1 + 2 + 3 + 100 \large =1+2+3+\cdots 100 = 100 ( 101 ) 2 \large =\dfrac{100(101)}{2} = 5050 \huge =\boxed{5050}


Equality occurs when: sin θ 1 1 = sin θ 2 2 = = sin θ 100 100 \dfrac{\sin \theta_1}{1}=\dfrac{\sin \theta_2}{\sqrt 2}=\cdots =\dfrac{\sin \theta _{100}}{\sqrt{100}}

Cool! \text{}

Harsh Shrivastava - 5 years, 2 months ago

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Thanks....... :-)

Rishabh Jain - 5 years, 2 months ago

Can you explain more this case of Cauchy-Schwartz-inequality, please? Thank you!

Anghel Corneliu - 5 years, 2 months ago
Adarsh Kumar
Mar 26, 2016

Just apply Cauchy inequality,for the numerator and the denominator will get cancelled,leaving 1 + 2 + 3 + . . . 100 = 5050 1+2+3+...100=5050 .

\text{} Correct.

Harsh Shrivastava - 5 years, 2 months ago

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Hehe!It looked tough but was pretty easy.

Adarsh Kumar - 5 years, 2 months ago

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