Trig Revolution 2020

Geometry Level 5

Over all angles A , B , C A, B, C of an acute triangle, find the minimum value of

2020 sin A sin B + sin C sin A 2020\sum \frac{\sqrt{\sin A}}{\sqrt{\sin B} + \sqrt{\sin C} - \sqrt{\sin A}}


The answer is 6060.

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1 solution

Priyanshu Mishra
Oct 16, 2016

By sine rule

sin A sin B + sin C sin A = a b + c a \huge\ \sum { \frac { \sqrt { \sin { A } } }{ \sqrt { \sin { B } } +\sqrt { \sin { C } } -\sqrt { \sin { A } } } =\sum { \frac { \sqrt { a } }{ \sqrt { b } +\sqrt { c } -\sqrt { a } } } } .

Now, let

x = b + c a y = c + a b z = a + b c \large\ {x=\sqrt { b } +\sqrt { c } -\sqrt { a } \\ y=\sqrt { c } +\sqrt { a } -\sqrt { b } \\ z=\sqrt { a } +\sqrt { b } -\sqrt { c }} .

Substituting it the inequality to be solved becomes

y + z 2 x \huge\ \sum { \frac { y+z }{ 2x } } , whose minimum value by AM-GM is 3 3 .

Thus,,

2020 sin A sin B + sin C sin A ( 2020 × 3 ) = 6060 \huge\ 2020\sum { \frac { \sqrt { \sin { A } } }{ \sqrt { \sin { B } } +\sqrt { \sin { C } } -\sqrt { \sin { A } } } } \ge \left( 2020\times 3 \right) =\boxed { 6060 }

This is a good start. However, your explanation has a slight gap, which is a common misconception made with inequalities.

  1. You have only shown that we have a lower bound on the value. In order for it to be minimum, this value must be attainable.

Keep writing more solutions and you will get the hang of this!

Calvin Lin Staff - 4 years, 8 months ago

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"you will get hang of this"

Sir i cannot understand by this line. What do you mean?

Should i show the equality case also?

Priyanshu Mishra - 4 years, 8 months ago

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It means that you will be able to write good solutions by keep writing more solutions over time.

Yes, you should show that the minimum value is attainable. i.e.: when A = B = C = 6 0 A=B=C = 60^\circ .

Pi Han Goh - 4 years, 8 months ago

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@Pi Han Goh Thanks . I got it.

Priyanshu Mishra - 4 years, 8 months ago

I agree that this has a good start, but your explanation has some gaps that need filling in:

[1] What motivates you to use the substitution x = b + c a , y = c + a b , z = a + b c x= \sqrt b + \sqrt c - \sqrt a, y = \sqrt c + \sqrt a - \sqrt b , z = \sqrt a + \sqrt b - \sqrt c in the first place? It appears that you have written a bunch of prefabricated steps which removes the mystery behind the problem.

[2] It is also worth clarifying why the denominator of the fraction in the given cyclic sum is never zero, otherwise, the sum can be undefined.

Pi Han Goh - 4 years, 8 months ago

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The substitution i did was to make the expression much simpler so that i can use AM-GM, Titu's lemma etc inequalities with ease.

I am working on my solution. I will modify it later.

Priyanshu Mishra - 4 years, 8 months ago

Try this also.

https://brilliant.org/problems/korea-proposal-to-imo/?ref_id=1266863

Priyanshu Mishra - 4 years, 8 months ago

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