x 5 + y 5 + z 5
Given that x , y and z are real numbers satisfying x + y + z = 0 and x 2 + y 2 + z 2 = 1 . And the maximum value of of the expression above can be expressed as A H K , where A , H and K are positive integers with K square-free and A , H copime. Find H + K + A .
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Question 1 : How did you make the conclusion that ∣ k ∣ ≤ 3 6 1 ? (Yes, I read your earlier sentence and I still couldn't make the connection)
Question 2 : How do you know that the maximum occurs at ( x , y , z ) = ( − 6 1 , − 6 1 , 6 2 ) ?
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The fact that f ( 1 / 6 ) = k − 1 / 3 6 ≤ 0 tells us that k ≤ 1 / 3 6 . The inequality about f ( − 1 / 6 ) does the rest. Where the function is maximized is a throwaway comment, and can be determined by inspection.
Alternatively, it is clear that S 5 is maximized when k = − 3 6 1 ; for this value of k , − 3 1 is a repeated root of the cubic f ( X ) ; it is easy to find the third root.
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Woahhhh why I didn't think of this?! You're so good! Thank you!
@MS HT , do you have any classical inequalities solution for this problem?
WLOG x ≤ y ≤ z . then, either x < 0 , y < 0 or y > 0 , z > 0 . If it were the latter, then S = x 5 + y 5 + z 5 would be negative. so, to maximise S ,we proceed with the first case, as S would be positve then.
let x = − a , y = − b with a , b > 0 we have a 2 + b 2 + ( a + b ) 2 = 1
S = ( a + b ) 5 − a 5 − b 5 = 5 a b 4 + 1 0 a 2 b 3 + 1 0 a 3 b 2 + 5 a 4 b = 5 a b ( a 3 + 2 a b ( a + b ) + b 3 ) = 5 a b ( a + b ) ( a 2 + a b + b 2 )
S = 2 5 a b ( a + b ) ≤ 8 5 ( a + b ) 3
we can use the fact that (it can be proved simply by expanding) 3 ( a + b ) 2 ≤ 4 ( a 2 + a b + b 2 )
So, ( a + b ) 3 ≤ ( 3 2 ) 2 3
Thus S ≤ 8 5 . ( 3 2 ) 2 3 = 3 6 5 6
Also, the equality occurs at x = y = − 6 1 , z = 6 2
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We have x y + x z + y z = 2 1 [ ( x + y + z ) 2 − ( x 2 + y 2 + z 2 ) ] = − 2 1 , and hence x , y , z are three real roots of the cubic equation f ( X ) = X 3 − 2 1 X + k = 0 for some constant k . The turning points of f ( X ) occur at X = ± 6 1 , and f ( ± 6 1 ) = k ∓ 3 6 1 . Since the cubic equation f ( X ) = 0 must have three real roots we must have f ( − 6 1 ) ≥ 0 ≥ f ( 6 1 ) , and hence ∣ k ∣ ≤ 3 6 1 .
If we define S n = x n + y n + z n then S n + 3 − 2 1 S n + 1 + k S n = 0 for all integers n . Thus S 5 = 2 1 S 3 − k S 2 = 2 1 ( 2 1 S 1 − k S 0 ) − k S 2 = − 2 5 k ≤ 6 6 5 = 3 6 5 6 , so that the answer is 5 + 6 + 3 6 = 4 7 . This maximum value is achieved when x = y = − 6 1 and z = 6 2 .