Trigequality

Algebra Level 5

Find the maximum value of k k such that a b c ( 2 a 2 b + 2 a c 2 + 2 b 2 c − k 2 a b c ) + a 4 c 2 + c 4 b 2 + b 4 a 2 ≥ 0 abc(2a^2b + 2ac^2 + 2b^2c - k^2abc) + a^4c^2 + c^4b^2 + b^4a^2 \ge 0 for all a , b , c a,b,c where a , b , c , k a,b,c,k are positive and real. Enter your answer as cos ⁡ ( π k ) \cos(\frac{\pi}{k}) Calculators are allowed.


The answer is 0.5.

Samuel Sturge by Samuel Sturge , 16, United Kingdom

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1 solution

Samuel Sturge
Jul 31, 2019

Expanding the brackets and rearranging yields 2 a 3 b 2 c + 2 c 3 a 2 b + 2 b 3 c 2 a + a 4 c 2 + c 4 b 2 + b 4 a 2 > = k 2 a 2 b 2 c 2 2a^3b^2c + 2c^3a^2b + 2b^3c^2a + a^4c^2 + c^4b^2 + b^4a^2 >= k^2a^2b^2c^2 Square rooting both sides: a 2 c + c 2 b + b 2 a > = k a b c a^2c + c^2b + b^2a >= kabc Dividing both sides by a b c abc and simplifying: a b + c a + b c > = k \frac{a}{b} + \frac {c}{a} + \frac {b}{c} >= k By the am - gm inequality: a b + c a + b c > = 3 \frac{a}{b} + \frac {c}{a} + \frac {b}{c} >= 3 . cos ⁡ ( π 3 ) \cos(\frac{\pi}{3}) = 0.5 0.5

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The problem is bad-written.

Andrea Bortolussi - 1 year, 10 months ago

I'm not sure if this holds generally,but: The symmetry of the equation in its variables implies that a=b=c for max. This gives the correct answer in all the questions of this type I have tried, including this one. Comments appreciated.

Ash Robson - 1 year, 9 months ago

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See wiki UVW substitution; it talks about Tehj's theorem which is pretty much your hypothesis.

Samuel Sturge - 1 year, 9 months ago

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