62208 and M

Algebra Level 3

62208 ( a 7 + b 7 + c 7 + d 7 ) 2 M ( a 2 + b 2 + c 2 + d 2 ) 7 62208\big(a^{7}+b^{7}+c^{7}+d^{7}\big)^{2}\le M\big(a^{2}+b^{2}+c^{2}+d^{2}\big)^{7}

What is the smallest positive integer M M such that this inequality holds true for all a , b , c , d a,b,c,d satisfying a , b , c , d R a,b,c,d\in \mathbb{R} and a + b + c + d = 0 ? a+b+c+d=0?


The answer is 8281.

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5 solutions

Kevin Tong
Dec 3, 2018

I was going to solve this conventionally, but then I noticed a nice "shortcut." I'm probably going to get a lot of criticism for this method, but oh well, it works. Since a + b + c + d = 0 a+b+c+d=0 , we can let d = a b c d=-a-b-c , so that 62208 ( a 7 + b 7 + c 7 + d 7 ) 2 = 2 8 3 5 ( a 7 + b 7 + c 7 ( a + b + c ) 7 ) 2 M ( a 2 + b 2 + c 2 + d 2 ) 7 = M ( a 2 + b 2 + c 2 + ( a + b + c ) 2 ) 7 62208(a^7+b^7+c^7+d^7)^2=2^8\cdot 3^5\cdot \left(a^7+b^7+c^7-(a+b+c)^7\right)^2\le M(a^2+b^2+c^2+d^2)^7=M\left(a^2+b^2+c^2+(a+b+c)^2\right)^7 2 8 3 5 ( a 7 + b 7 + c 7 ( a + b + c ) 7 ) 2 M ( a 2 + b 2 + c 2 + ( a + b + c ) 2 ) 7 2^8\cdot 3^5\cdot\left(a^7+b^7+c^7-(a+b+c)^7\right)^2\le M\left(a^2+b^2+c^2+(a+b+c)^2\right)^7 f ( a , b , c ) = M ( a 2 + b 2 + c 2 + ( a + b + c ) 2 ) 7 2 8 3 5 ( a 7 + b 7 + c 7 ( a + b + c ) 7 ) 2 0 f(a,b,c)=M\left(a^2+b^2+c^2+(a+b+c)^2\right)^7-2^8\cdot 3^5\cdot\left(a^7+b^7+c^7-(a+b+c)^7\right)^2\ge 0 Notice that f ( a , b , c ) f(a,b,c) is symmetric, meaning that f ( a , b , c ) = f ( b , a , c ) = f ( c , b , a ) = f(a,b,c)=f(b,a,c)=f(c,b,a)=\cdots for all permutations of ( a , b , c ) (a,b,c) . If we consider optimizing f ( a , b , c ) f(a,b,c) by taking the partial derivative with respect to each of the variables, we would get a symmetric equation, which would imply that a = b = c a=b=c when the function is optimized. Thus, we obtain 2 8 3 5 ( 3 a 7 ( 3 a ) 7 ) 2 = M ( 3 a 2 + ( 3 a ) 2 ) 7 2^8\cdot 3^5\cdot \left(3a^7-(3a)^7\right)^2=M\left(3a^2+(3a)^2\right)^7 2 8 3 7 ( 1 3 6 ) 2 a 14 = 1 2 7 M a 14 = 2 14 3 7 M a 14 2^8\cdot 3^7\cdot \left(1-3^6\right)^2\cdot a^{14}=12^7\cdot M\cdot a^{14}=2^{14}\cdot 3^7\cdot M\cdot a^{14} M = 2 8 3 7 ( 3 6 1 ) 2 2 14 3 7 = ( ( 3 3 1 ) ( 3 3 + 1 ) ) 2 2 6 = 1 3 2 7 2 = 8281 M=\frac{2^8\cdot 3^7\cdot (3^6-1)^2}{2^{14}\cdot 3^7}=\frac{((3^3-1)(3^3+1))^2}{2^6}=13^2\cdot 7^2=\boxed{8281}

There will be a turning point at a = b = c a=b=c . Why must it be a (global) maximum?

Mark Hennings - 2 years, 6 months ago

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My initial thoughts on this was that since the function would be optimized in all 3 directions (in the a , b , a, b, and c c direction), the point at a = b = c a=b=c should be optimal. The reasoning may be flawed (and I'm sure it is), but it's pretty intuitive.

Kevin Tong - 2 years, 6 months ago

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When you use a word like "optimize" you end up thinking there is only one option. There are two choices to choose between, "maximize" or "minimize". If if it were reasonable to think that a = b = c a=b=c would be a local maximum, you have not thought about whether it provides a global maximum.

It's very easy just to plug in a = b = c a=b=c to these problems, but doing so does not prove the result, even if it does get the correct answer.

Mark Hennings - 2 years, 6 months ago

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@Mark Hennings I am certain that one day someone will characterize precisely when a function is "nice enough" to extremize it by setting all variables to be equal, and then smart kids will be able to cheese high-school math competition problems by referring to that theorem.

Vilim Lendvaj - 2 years, 6 months ago

But if we take a= b =c =d =0 than M will be 1

TRIAMBIKA GARG - 2 years, 2 months ago

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True, but M = 1 M=1 will not work for all reals a , b , c , d a,b,c,d with a + b + c + d = 0 a+b+c+d=0 .

Mark Hennings - 2 years, 2 months ago
Mark Hennings
Jun 17, 2018

Since the problem is homogeneous, we need to extremize F = a 7 + b 7 + c 7 + d 7 F = a^7 + b^7 + c^7 + d^7 subject to the constraints a + b + c + d = 0 a+b+c+d = 0 and a 2 + b 2 + c 2 + d 2 = 1 a^2 + b^2 + c^2 + d^2 = 1 . Putting d = ( a + b + c ) d = -(a+b+c) , this is the same as extremizing F = a 7 + b 7 + c 7 ( a + b + c ) 7 F = a^7 + b^7 + c^7 - (a+b+c)^7 subject to the constraint a 2 + b 2 + c 2 + ( a + b + c ) 2 = 1 a^2 + b^2 + c^2 + (a+b+c)^2 = 1 .

If we now put X = b + c X = b+c , Y = a + c Y = a+c , Z = a + b Z = a+b , this is the same as extremizing F = 2 7 [ ( Y + Z X ) 7 + ( X + Z Y ) 7 + ( X + Y Z ) 7 ( X + Y + Z ) 7 ] = 7 16 X Y Z [ 3 ( X 4 + Y 4 + Z 4 ) + 10 ( X 2 Y 2 + X 2 Z 2 + Y 2 Z 2 ) ] = 7 16 X Y Z [ 5 ( X 2 + Y 2 + Z 2 ) 2 2 ( X 4 + Y 4 + Z 4 ) ] \begin{aligned} F & = \; 2^{-7}\big[(Y+Z-X)^7 + (X+Z-Y)^7 + (X+Y-Z)^7 - (X+Y+Z)^7\big] \\ & = \; -\tfrac{7}{16}XYZ\big[3(X^4 + Y^4 + Z^4) + 10(X^2Y^2 + X^2Z^2 + Y^2Z^2)\big] \; = \; -\tfrac{7}{16}XYZ\big[5(X^2 + Y^2 + Z^2)^2 - 2(X^4 + Y^4 + Z^4)\big] \end{aligned} subject to the constraint X 2 + Y 2 + Z 2 = 1 X^2 + Y^2 + Z^2 = 1 . Thus we want to extremize F = 7 16 X Y Z [ 5 2 ( X 4 + Y 4 + Z 4 ) ] F \; = \; -\tfrac{7}{16}XYZ\big[5 - 2(X^4 + Y^4 + Z^4)\big] subject to the constraint X 2 + Y 2 + Z 2 = 1 X^2 + Y^2 + Z^2 = 1 .

The Cauchy-Schwarz Inequality tells us that ( X 2 + Y 2 + Z 2 ) 2 3 ( X 4 + Y 4 + Z 4 ) (X^2 + Y^2 + Z^2)^2 \le 3(X^4 + Y^4 + Z^4) and so X 4 + Y 4 + Z 4 1 3 X^4 + Y^4 + Z^4 \ge \tfrac13 provided that X 2 + Y 2 + Z 2 = 1 X^2 + Y^2 + Z^2 = 1 (with equality when X = Y = Z = 1 3 |X| = |Y| = |Z| = \tfrac{1}{\sqrt{3}} ). The AM/GM Inequality tells us that X Y Z 2 3 1 3 ( X 2 + Y 2 + Z 2 ) |XYZ|^{\frac23} \le \tfrac13(X^2 + Y^2 + Z^2) , so that X Y Z 1 3 3 |XYZ| \le \tfrac{1}{3\sqrt{3}} provided that X 2 + Y 2 + Z 2 = 1 X^2 + Y^2 + Z^2 = 1 (with equality when X = Y = Z = 1 3 |X| = |Y| = |Z| = \tfrac{1}{\sqrt{3}} ). Thus, subject to the constraint X 2 + Y 2 + Z 2 = 1 X^2 + Y^2 + Z^2 = 1 , we have F 7 16 × 1 3 3 × [ 5 2 3 ] = 91 144 3 |F| \le \tfrac{7}{16} \times \tfrac{1}{3\sqrt{3}} \times \big[5 - \tfrac23\big] \; = \; \tfrac{91}{144\sqrt{3}} with equality when X = Y = Z = 1 3 |X| = |Y| = |Z| = \tfrac{1}{\sqrt{3}} .

Thus we have shown that ( a 7 + b 7 + c 7 + d 7 ) 2 ( 91 144 3 ) 2 ( a 2 + b 2 + c 2 + d 2 ) 7 = 8281 62208 ( a 2 + b 2 + c 2 + d 2 ) 7 \big(a^7 + b^7 + c^7 + d^7\big)^2 \; \le \; \left(\tfrac{91}{144\sqrt{3}}\right)^2(a^2 + b^2 + c^2 + d^2)^7 \; = \; \tfrac{8281}{62208}(a^2 + b^2 + c^2 + d^2)^7 provided that a + b + c + d = 0 a + b + c + d = 0 , with equality when (for example) a = b = c = 1 2 3 a = b = c = \tfrac{1}{2\sqrt{3}} and d = 1 2 3 d = -\tfrac12\sqrt{3} , and so we deduce that M = 8281 M = \boxed{8281} .

Wow!Great. Using X,Y,Z to substitute for b+c,c+a,a+b simplifies the final expression effectively.But somehow I missed my good luck.

In fact,I spent a lot of time before getting the identity

a 7 + b 7 + c 7 ( a + b + c ) 7 = 7 a^{7}+b^{7}+c^{7}-(a+b+c)^{7}=-7 ( Π ( a + b ) ) \color{#20A900}(\Pi(a+b)) ( ( a 2 + a b ) 2 + a b c a ) \Big((\sum a^{2}+\sum ab)^{2}+abc\sum a\Big)

(To do that,first note that when a = b a=-b ,the expression is equal to 0 0 ,so it must have a factor ( a + b ) (a+b) ,and similarly, ( b + c ) (b+c) and ( c + a ) (c+a) .)

Haosen Chen - 2 years, 11 months ago

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it is a symmetric expression. We are exploiting that fact. After factorising guessing the substitution is not clear. x=a+b,... Is making things so simple. Only reason to choose that substitution is because we have factors a+b etc. How to arrive at this and why things are becoming simple.

Srikanth Tupurani - 2 years, 7 months ago

This is quite above my level, but I'm trying to understand: why can we pick the second constraint (the sum of squares equals 1)? Thanks

Laszlo Kocsis - 2 years, 6 months ago

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The expression is homogeneous, in the sense that if a , b , c , d a,b,c,d satisfies the inequality, then so does u a , u b , u c , u d ua,ub,uc,ud for any u u . Thus it is enough to restrict attention to those solutions for which (for example) a 2 + b 2 + c 2 + d 2 a^2+b^2+c^2+d^2 takes a particular value, such as 1 1 .

Mark Hennings - 2 years, 6 months ago

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Thanks, now I understand.

Laszlo Kocsis - 2 years, 6 months ago

what about a = b = 1, c= d= -1 ? Then a+b+c+d = 0, and we find 0 < M * 4**7 ? The answer could be M = 1 ?

michel gelders - 2 years, 6 months ago

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M = 8281 M=8281 is the smallest value such that the inequality holds for all a , b , c , d a,b,c,d with a + b + c + d = 0 a+b+c+d=0 . There are certainly some values of a , b , c , d a,b,c,d for which a smaller value of M M would work, but M = 1 M=1 (for example), would not work with a = b = c = 1 a=b=c=1 , d = 3 d=-3

Mark Hennings - 2 years, 6 months ago

I simply chose a = b = c = 1 / 3 a=b=c=1/3 and d = 1 d=1 , and proceeded to solve it numerically from there, producting the answer 8291 \boxed{8291} .

Bert Seegmiller - 2 years, 5 months ago

... I see I can't edit comments, once submitted, neither preview them.

I need to correct a couple of things in my original comment: * 1 / 3 1/3 should be 1 / 3 -1/3 * 8291 should be 8281

Bert Seegmiller - 2 years, 5 months ago
Laurent Shorts
Dec 6, 2018

M 6 2 208 \dfrac{M}{62'208} is a maximum of f ( a , b , c , d ) = ( a 7 + b 7 + c 7 + d 7 ) 2 ( a 2 + b 2 + c 2 + d 2 ) 7 f(a,b,c,d)=\dfrac{(a^7+b^7+c^7+d^7)^2}{(a^2+b^2+c^2+d^2)^7} , when g ( a , b , c , d ) = a + b + c + d = 0 g(a,b,c,d)=a+b+c+d=0 .

Extrema occur when f \nabla f and g \nabla g are collinear (where f \nabla f is the vector of the partial derivatives of f f ).

With S = ( a 7 + b 7 + c 7 + d 7 ) S=(a^7+b^7+c^7+d^7) and T = ( a 2 + b 2 + c 2 + d 2 ) T=(a^2+b^2+c^2+d^2) , we have g = ( 1 ; 1 ; 1 ; 1 ) \nabla g=(1; 1; 1; 1) and

f = 14 S T 6 ( T a 6 S a ; T b 6 S b ; T c 6 S c ; T d 6 S d ) \nabla f = 14·S·T^6\big(T·a^6 - S·a\ ; \ \ T·b^6 - S·b\ ; \ \ T·c^6 - S·c\ ; \ \ T·d^6 - S·d \big) .

We have collinearity:

  • when S = 0 S=0 (which gives M = 0 M=0 for equality, that is clearly a local minimum and not a maximum);

  • when T = 0 T=0 (which gives a = b = c = d = 0 a=b=c=d=0 and then M M can be anything for equality);

  • when T x 6 S x = T y 6 S y T·x^6-S·x\ =\ T·y^6-S·y , for x x and y y any of a a , b b , c c or d d .

T x 6 S x = T y 6 S y x = y T·x^6-S·x\ =\ T·y^6-S·y\ \Leftrightarrow\ x=y or x 6 y 6 x y = S T \dfrac{x^6-y^6}{x-y}=\dfrac{S}{T} .

If x 6 y 6 x y = S T \dfrac{x^6-y^6}{x-y}=\dfrac ST and x 6 z 6 x z = S T \dfrac{x^6-z^6}{x-z}=\dfrac ST , then z = y z=y (because h ( y ) = x 6 y 6 x y h(y)=\dfrac{x^6-y^6}{x-y} is always increasing, therefore bijective after checking what occurs at y = x y=x ).

Therefore, a a , b b , c c and d d must be the same value x x , or be of two distinct values x x and y y .

If all are the same values, then the sum being 0 implies that all are 0 and that T T =0 (case already seen).

If values are x , x , y , y {x,x,y,y} , then the sum being 0 implies that x = y x=-y and therefore S = 0 S=0 (case already seen).

If values are x , x , x , y {x,x,x,y} , then the sum being 0 implies that y = 3 x y=-3x .

In this extreme case, we have: M 6 2 208 = f ( x , x , x , 3 x ) = ( 3 3 7 ) 2 x 14 ( 3 + ( 3 ) 2 ) 7 x 14 = 2 18 4 2 1 2 7 \dfrac{M}{62'208}=f(x,x,x,-3x)=\dfrac{(3-3^7)^2·x^{14}}{(3+(-3)^2)^7·x^{14}}=\dfrac{2'184^2}{12^7}

That is, M = 6 2 208 2 18 4 2 1 2 7 = 2 8 3 5 2 6 3 2 7 2 1 3 2 2 14 3 7 = 7 2 1 3 2 = 8 281 M=62'208·\dfrac{2'184^2}{12^7}=2^8·3^5·\dfrac{2^6·3^2·7^2·13^2}{2^{14}·3^7}=7^2·13^2=\boxed{8'281}

I agree with you .But if we take the case of a=1,b=c=(-1/2) and d=0 ,then M will be 60,293.90.So the function (a^7+b^7+c^7+d^7)²/(a²+b²+c²+d²)^7 has a maximum value very near to 1.

Alapan Das - 2 years, 6 months ago

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I find f ( 1 , 1 2 , 1 2 , 0 ) = 0.056713 f(1,-\frac12,-\frac12,0)=0.056713 and M = 3 , 528 M=3,528 . Am I wrong?

Laurent Shorts - 2 years, 6 months ago
K T
Dec 9, 2018

Not a proof, but a fruitful thought. We want to maximize the fraction f = ( a 7 + . . ) 2 ( a 2 + . . ) 7 f = \frac{(a^7+..)^2}{(a^2+..)^7} and we observe that

  • with only 2 terms a and b, they would have to satisfy b=-a and the fraction would be 0
  • when adding more terms, the fraction is the highest when one of them (say a) is different while the others are equal
  • multiplying a, b, c and d by the same factor will not change f, so we will set a=3 and b=c=d=-1.

In our case we get M = 62208 f m a x = 62208 × ( 3 7 3 ) 2 ( 3 2 + 3 ) 7 = 8281 M=62208 f_{max} = 62208×\frac{(3^7-3)^2}{(3^2+3)^7}=8281

I agree with you .But if we take the case of a=1,b=c=(-1/2) and d=0 ,then M will be 60,293.90.So the function (a^7+b^7+c^7+d^7)²/(a²+b²+c²+d²)^7 has a maximum value very near to 1.

Alapan Das - 2 years, 6 months ago
Vinod Kumar
Dec 5, 2018

Assume a=b=c with a+b+c+d=0 and obtain the following desired result:

62208×(1^7 + 1^7 + 1^7 + (-3)^7)^2/(1^2 + 1^2 + 1^2 + (-3)^2)^7=8281

Answer=8281

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