3000 Follower Problem 1

Algebra Level 5

{ F 2 + A 2 + N 2 = 3 min ( F + A , A + N , F + N ) > 2 F ( A + N F ) 2 + A ( N + F A ) 2 + N ( F + A N ) 2 S 3000 ( F A N ) 2 \begin{cases} F^2+A^2+N^2 =3 \\ \min \left( F+A, A+N, F+N \right) > \sqrt 2 \\ \frac{F}{\left( A+N-F \right)^2}+\frac{A}{\left( N+F-A \right)^2}+\frac{N}{\left(F+A-N \right)^2} \geq \frac{S}{3000 \left( FAN \right)^2} \end{cases}

For what maximum value of S S will the above conditions be always met simultaneously for positive real numbers F , A F,A and N N ?

Clarification:

  • F A N = F × A × N FAN= F \times A \times N .


The answer is 9000.0.

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1 solution

Shridam Mahajan
Feb 21, 2016

For Mcq lovers it's a great and interesting problem. Just assume f=a=n=1 as it would give the max. Value and also there's an interesting method to solve it

You got lucky! This assumption doesn't work always. For such cases you can refer to Inequalities with strange equality conditions .

Can you give the general solution of the problem?

Sandeep Bhardwaj - 5 years, 3 months ago

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I solved it at 110 points ._.

Why does this always happen to me?

Mehul Arora - 5 years, 3 months ago

Yes sir I got lucky!! But my method involves two inequalities. I have only solved it for this particular problem. Not able to generalise it :(

Shridam Mahajan - 5 years, 3 months ago

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