Problem 19

Algebra Level 5

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

Given that x , y x,y and z z are real numbers satisfying x + y + z = 0 x+y+z=0 and x 2 + y 2 + z 2 = 1 x^2+y^2+z^2 = 1 . And the maximum value of of the expression above can be expressed as H K A \dfrac{ H\sqrt K}A , where A , H A,H and K K are positive integers with K K square-free and A , H A,H copime. Find H + K + A H+K+A .


This problem part of this set: It's all about the inequality .
Excerpted from the 2012 college, university entrance examination for students in B (Maths, Chemistry, Biology)


The answer is 47.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Mark Hennings
Feb 28, 2016

We have x y + x z + y z = 1 2 [ ( x + y + z ) 2 ( x 2 + y 2 + z 2 ) ] = 1 2 xy + xz + yz \,=\, \tfrac12\big[(x + y + z)^2 - (x^2 + y^2 + z^2)\big] \,=\, -\tfrac12 , and hence x , y , z x,y,z are three real roots of the cubic equation f ( X ) = X 3 1 2 X + k = 0 f(X) \; = \; X^3 - \tfrac12X + k \; = \; 0 for some constant k k . The turning points of f ( X ) f(X) occur at X = ± 1 6 X = \pm \tfrac{1}{\sqrt{6}} , and f ( ± 1 6 ) = k 1 3 6 f\big(\pm\tfrac{1}{\sqrt{6}}\big) \,=\, k \mp \tfrac{1}{3\sqrt{6}} . Since the cubic equation f ( X ) = 0 f(X) = 0 must have three real roots we must have f ( 1 6 ) 0 f ( 1 6 ) f\big(-\tfrac{1}{\sqrt{6}}\big) \,\ge \, 0 \,\ge\, f\big(\tfrac{1}{\sqrt{6}}\big) , and hence k 1 3 6 |k| \le \tfrac{1}{3\sqrt{6}} .

If we define S n = x n + y n + z n S_n \,=\, x^n + y^n + z^n then S n + 3 1 2 S n + 1 + k S n = 0 S_{n+3} - \tfrac12S_{n+1} + kS_n \,=\, 0 for all integers n n . Thus S 5 = 1 2 S 3 k S 2 = 1 2 ( 1 2 S 1 k S 0 ) k S 2 = 5 2 k 5 6 6 = 5 6 36 , S_5 \; = \; \tfrac12S_3 - kS_2 \; = \; \tfrac12\big(\tfrac12S_1 - kS_0\big) - kS_2 \; = \; -\tfrac52k \; \le \; \tfrac{5}{6\sqrt{6}} \; = \; \tfrac{5\sqrt{6}}{36}\;, so that the answer is 5 + 6 + 36 = 47 5+6+36 = \boxed{47} . This maximum value is achieved when x = y = 1 6 x=y=-\tfrac{1}{\sqrt{6}} and z = 2 6 z = \tfrac{2}{\sqrt{6}} .

Question 1 : How did you make the conclusion that k 1 3 6 |k| \leq \dfrac1{3\sqrt6} ? (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 ) = ( 1 6 , 1 6 , 2 6 ) (x,y,z) = \left( -\dfrac1{\sqrt6}, -\dfrac1{\sqrt6} , \dfrac2{\sqrt6} \right) ?

Pi Han Goh - 5 years, 2 months ago

Log in to reply

The fact that f ( 1 / 6 ) = k 1 / 3 6 0 f(1/\sqrt{6}) = k - 1/3\sqrt{6} \le 0 tells us that k 1 / 3 6 k \le 1/3\sqrt{6} . The inequality about f ( 1 / 6 ) f(-1/\sqrt{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 S_5 is maximized when k = 1 3 6 k = -\tfrac{1}{3\sqrt{6}} ; for this value of k k , 1 3 -\tfrac{1}{\sqrt{3}} is a repeated root of the cubic f ( X ) f(X) ; it is easy to find the third root.

Mark Hennings - 5 years, 2 months ago

Log in to reply

Woahhhh why I didn't think of this?! You're so good! Thank you!

Pi Han Goh - 5 years, 2 months ago

@MS HT , do you have any classical inequalities solution for this problem?

Pi Han Goh - 5 years, 2 months ago
Rohit Kumar
Mar 2, 2016

WLOG x y z x \leq y \leq z . then, either x < 0 , y < 0 x < 0 , y < 0 or y > 0 , z > 0 y > 0 , z >0 . If it were the latter, then S = x 5 + y 5 + z 5 S = x^5 + y^5 + z^5 would be negative. so, to maximise S S ,we proceed with the first case, as S S would be positve then.

let x = a , y = b x = -a, y = -b with a , b > 0 a,b>0 we have a 2 + b 2 + ( a + b ) 2 = 1 a^2 + b^2 +(a + b)^2 = 1

S = ( a + b ) 5 a 5 b 5 = 5 a b 4 + 10 a 2 b 3 + 10 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 = (a+b)^5 - a^5 -b^5 = 5ab^4 + 10a^2b^3 + 10a^3b^2 + 5a^4b = 5ab(a^3 + 2ab(a+b) + b^3) = 5ab(a+b)(a^2 + ab + b^2)

S = 5 2 a b ( a + b ) 5 8 ( a + b ) 3 S = \dfrac{5}{2}ab(a+b) \leq \dfrac{5}{8}(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 ) 3(a+b)^2 \leq 4(a^2 + ab + b^2)

So, ( a + b ) 3 ( 2 3 ) 3 2 (a+b)^3 \leq (\dfrac{2}{3})^{\frac{3}{2}}

Thus S 5 8 . ( 2 3 ) 3 2 = 5 6 36 S \leq \dfrac{5}{8}.(\dfrac{2}{3})^{\frac{3}{2}} = \boxed{\dfrac{5\sqrt{6}}{36}}

Also, the equality occurs at x = y = 1 6 , z = 2 6 x = y = -\dfrac{1}{\sqrt{6}} , z = \dfrac{2}{\sqrt{6}}

Nikola Djuric
Feb 29, 2016

x=y,so z=-2x

x^2+x^2+4x^2=1,so x^2=1/6

x^5+x^5-32x^5=-30x^5 so for x=-1/√6 (y=-1/√6,z=2/√6)

max is 30/(36√6)=5/(6√6)=5√6/36

so 5+6+36=47

Why must we have x = y x = y ?

Calvin Lin Staff - 5 years, 3 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...