AM-GM Won't Work Here

Algebra Level 5

x 5 y + y 5 z + z 5 x \large x^5 y + y^5 z + z^5 x

Let x , y , x, y, and z z be non-negative reals such that x + y + z = 1 x+y+z=1 .

The maximum value of the above expression can be represented as a b c d \dfrac {a^b}{c^d} , where a a and c c are not perfect powers, and a , b , c , d a,b,c,d are positive integers. Find the value of a + b + c + d a+b+c+d .

Bonus: Generalize this for the expression x n y + y n z + z n x x^n y + y^n z + z^n x .


The answer is 22.

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2 solutions

Mark Hennings
Oct 30, 2016

For any n 2 n \ge 2 , let f n ( x , y , z ) = x n y + y n z + z n x f_n(x,y,z) = x^ny + y^nz + z^nx . We want to maximize f n ( x , y , z ) f_n(x,y,z) over all nonnegative x , y , z x,y,z with x + y + z = 1 x+y+z=1 . Without loss of generality, we can assume that x y , z x \ge 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 1 1 x n y n z n x y z = 0 0 1 x n z n y n z n z n x z y z 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 \begin{array}{rcl} \displaystyle f_n(x,y,z) - f_n(x,z,y) & = & x^ny + y^nz + z^nx - x^nz - z^ny - y^nx \\ & = & \displaystyle \left| \begin{array}{ccc} 1 & 1 & 1 \\ x^n & y^n & z^n \\ x & y & z \end{array}\right| \; = \; \left|\begin{array}{ccc} 0 & 0 & 1 \\ x^n - z^n & y^n - z^n & z^n \\ x - z & y - z & z \end{array} \right| \\ & = & \displaystyle (x - z)(y - z)\left[ \sum_{j=0}^{n-1}x^j z^{n-1-j} - \sum_{j=0}^{n-1} y^j z^{n-1-j}\right] \\ & = & \displaystyle (x - z)(y - z) \sum_{j=0}^{n-1} \big(x^j - y^j\big)z^{n-1-j} \\ & = & \displaystyle (x - z)(y - z)(x - y) \sum_{j=1}^{n-1} \sum_{k=0}^{j-1} x^k y^{j-1-k} z^{n-1-j} \end{array} 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 x,y,z of degree n 2 n-2 ). Thus the largest possible value of f n ( x , y , z ) f_n(x,y,z) will occur when x y z x \ge y \ge 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 \begin{array}{rcl} f_n(x+z,y,0) - f_n(x,y,z) & = & (x+z)^ny - x^ny - y^nz - z^nx \; \ge \; x^ny + nx^{n-1}yz - x^ny - y^nz - z^nx \\ & = & (x^{n-1} - y^{n-1})yz + \big[(n-1)x^{n-2}y - z^{n-1}\big]xz \; \ge \; 0 \end{array} so the largest possible value of f n ( x , y , z ) f_n(x,y,z) occurs when when x y x \ge y and z = 0 z=0 .

Thus we simply need to maximize f n ( x , y , 0 ) = x n y f_n(x,y,0) = x^ny subject to the conditions x , y 0 x,y \ge 0 and x + y = 1 x+y=1 . Considering the arithmetic and geometric means of n n copies of 1 n x \tfrac1nx and one copy of y y , we deduce that ( x n y n n ) 1 n + 1 1 n + 1 ( n × 1 n x + y ) = 1 n + 1 ( x + y ) = 1 n + 1 \left(\frac{x^ny}{n^n}\right)^{\frac1{n+1}} \; \le \; \tfrac1{n+1}\big(n \times \tfrac1nx + y\big) \; = \; \tfrac1{n+1}(x+y) \; = \; \tfrac1{n+1} and hence, subject to the condition x + y = 1 x+y = 1 , f ( x , y , 0 ) = x n y n n ( n + 1 ) n + 1 f(x,y,0) \; = \; x^ny \; \le \; \frac{n^n}{(n+1)^{n+1}} We note that equality occurs when x = n n + 1 x = \tfrac{n}{n+1} , y = 1 n + 1 y = \tfrac1{n+1} . Thus the maximum value of f n ( x , y , z ) f_n(x,y,z) over all x , y , z 0 x,y,z \ge 0 with x + y + z = 1 x+y+z=1 is f ( n n + 1 , 1 n + 1 , 0 ) = n n ( n + 1 ) n + 1 f(\tfrac{n}{n+1},\tfrac1{n+1},0) = \frac{n^n}{(n+1)^{n+1}} .

Putting n = 5 n=5 , the maximum value for f 5 ( x , y , z ) f_5(x,y,z) , subject to the constraints, is 5 5 6 6 \frac{5^5}{6^6} , which makes the answer to the question 5 + 5 + 6 + 6 = 22 5+5+6+6 = \boxed{22} .

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.

Calvin Lin Staff - 4 years, 7 months ago

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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 3\times3 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 n .

Mark Hennings - 4 years, 7 months ago
P C
Dec 12, 2016

Here's the AM - GM solution, first I'll solve the generalized problem. From the condition, we concluded 0 x , y , z 1 0\leq x,y,z\leq 1 . For any real 0 a 1 0\leq a\leq 1 , we always get a n a , n Z a^n\leq a, n\in\mathbb{Z} . If I set z = m i n { x , y , z } z=min\{x,y,z\} , then x n y + y n z + z n x x n y + y n z + z x x^ny+y^nz+z^nx\leq x^ny+y^nz+zx The equality holds when z = 0 z=0 , so now x n y x^ny is what we have left in the R H S RHS , the condition also changed to x + y = 1 x+y=1 . 1 = x n + x n + . . . + x n + y A M G M ( n + 1 ) x n y n n n + 1 1=\frac{x}{n}+\frac{x}{n}+...+\frac{x}{n}+y\stackrel{AM-GM}\geq (n+1)\sqrt[n+1]{\frac{x^ny}{n^n}} x n y n n ( n + 1 ) n + 1 \therefore x^ny\leq\frac{n^n}{(n+1)^{n+1}} The equality holds when ( x , y , z ) = ( n n + 1 , 1 n + 1 , 0 ) (x,y,z)=\big(\frac{n}{n+1},\frac{1}{n+1},0\big) and its permutations.

Getting back to the main problem, we see n = 5 n=5 so the maximum is 5 5 6 6 \frac{5^5}{6^6} and the answer is 22 22

Note : In the application of AM-GM, x n \frac{x}{n} is repeated n n times

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