x 5 y + y 5 z + z 5 x
Let x , y , and z be non-negative reals such that x + y + z = 1 .
The maximum value of the above expression can be represented as c d a b , where a and c are not perfect powers, and a , b , c , d are positive integers. Find the value of a + b + c + d .
Bonus: Generalize this for the expression x n y + y n z + z n x .
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Very nice approach with smoothing.
For the factorization, as done here , I found it easier to prove it "after the fact", then trying to explain how one arrived at it.
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That, of course, assumes that you have come up with the idea of what the "sum of monomials" factor actually is. Doing column operations on a 3 × 3 determinant, which is very similar to a Vandermonde determinant (for which this technique is known to give the answer), is pretty straightforward, even in the case of general n .
Here's the AM - GM solution, first I'll solve the generalized problem. From the condition, we concluded 0 ≤ x , y , z ≤ 1 . For any real 0 ≤ a ≤ 1 , we always get a n ≤ a , n ∈ Z . If I set z = m i n { x , y , z } , then x n y + y n z + z n x ≤ x n y + y n z + z x The equality holds when z = 0 , so now x n y is what we have left in the R H S , the condition also changed to x + y = 1 . 1 = n x + n x + . . . + n x + y ≥ A M − G M ( n + 1 ) n + 1 n n x n y ∴ x n y ≤ ( n + 1 ) n + 1 n n The equality holds when ( x , y , z ) = ( n + 1 n , n + 1 1 , 0 ) and its permutations.
Getting back to the main problem, we see n = 5 so the maximum is 6 6 5 5 and the answer is 2 2
Note : In the application of AM-GM, n x is repeated n times
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For any n ≥ 2 , let f n ( x , y , z ) = x n y + y n z + z n x . We want to maximize f n ( x , y , z ) over all nonnegative x , y , z with x + y + z = 1 . Without loss of generality, we can assume that x ≥ y , z . Then f n ( x , y , z ) − f n ( x , z , y ) = = = = = x n y + y n z + z n x − x n z − z n y − y n x ∣ ∣ ∣ ∣ ∣ ∣ 1 x n x 1 y n y 1 z n z ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ ∣ ∣ 0 x n − z n x − z 0 y n − z n y − z 1 z n z ∣ ∣ ∣ ∣ ∣ ∣ ( x − z ) ( y − z ) [ j = 0 ∑ n − 1 x j z n − 1 − j − j = 0 ∑ n − 1 y j z n − 1 − j ] ( x − z ) ( y − z ) j = 0 ∑ n − 1 ( x j − y j ) z n − 1 − j ( x − z ) ( y − z ) ( x − y ) j = 1 ∑ n − 1 k = 0 ∑ j − 1 x k y j − 1 − k z n − 1 − j and we note that this last double sum is a sum of nonnegative terms (it is the sum of all the monomials in x , y , z of degree n − 2 ). Thus the largest possible value of f n ( x , y , z ) will occur when x ≥ y ≥ z . But then f n ( x + z , y , 0 ) − f n ( x , y , z ) = = ( x + z ) n y − x n y − y n z − z n x ≥ x n y + n x n − 1 y z − x n y − y n z − z n x ( x n − 1 − y n − 1 ) y z + [ ( n − 1 ) x n − 2 y − z n − 1 ] x z ≥ 0 so the largest possible value of f n ( x , y , z ) occurs when when x ≥ y and z = 0 .
Thus we simply need to maximize f n ( x , y , 0 ) = x n y subject to the conditions x , y ≥ 0 and x + y = 1 . Considering the arithmetic and geometric means of n copies of n 1 x and one copy of y , we deduce that ( n n x n y ) n + 1 1 ≤ n + 1 1 ( n × n 1 x + y ) = n + 1 1 ( x + y ) = n + 1 1 and hence, subject to the condition x + y = 1 , f ( x , y , 0 ) = x n y ≤ ( n + 1 ) n + 1 n n We note that equality occurs when x = n + 1 n , y = n + 1 1 . Thus the maximum value of f n ( x , y , z ) over all x , y , z ≥ 0 with x + y + z = 1 is f ( n + 1 n , n + 1 1 , 0 ) = ( n + 1 ) n + 1 n n .
Putting n = 5 , the maximum value for f 5 ( x , y , z ) , subject to the constraints, is 6 6 5 5 , which makes the answer to the question 5 + 5 + 6 + 6 = 2 2 .