6 2 2 0 8 ( a 7 + b 7 + c 7 + d 7 ) 2 ≤ M ( a 2 + b 2 + c 2 + d 2 ) 7
What is the smallest positive integer M such that this inequality holds true for all a , b , c , d satisfying a , b , c , d ∈ R and a + b + c + d = 0 ?
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There will be a turning point at a = b = c . Why must it be a (global) maximum?
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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 , and c direction), the point at a = b = c should be optimal. The reasoning may be flawed (and I'm sure it is), but it's pretty intuitive.
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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 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 to these problems, but doing so does not prove the result, even if it does get the correct answer.
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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.
But if we take a= b =c =d =0 than M will be 1
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True, but M = 1 will not work for all reals a , b , c , d with a + b + c + d = 0 .
Since the problem is homogeneous, we need to extremize F = a 7 + b 7 + c 7 + d 7 subject to the constraints a + b + c + d = 0 and a 2 + b 2 + c 2 + d 2 = 1 . Putting d = − ( a + b + c ) , this is the same as extremizing 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 .
If we now put X = b + c , Y = a + c , 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 ] = − 1 6 7 X Y Z [ 3 ( X 4 + Y 4 + Z 4 ) + 1 0 ( X 2 Y 2 + X 2 Z 2 + Y 2 Z 2 ) ] = − 1 6 7 X Y Z [ 5 ( X 2 + Y 2 + Z 2 ) 2 − 2 ( X 4 + Y 4 + Z 4 ) ] subject to the constraint X 2 + Y 2 + Z 2 = 1 . Thus we want to extremize F = − 1 6 7 X Y Z [ 5 − 2 ( X 4 + Y 4 + Z 4 ) ] subject to the constraint 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 ) and so X 4 + Y 4 + Z 4 ≥ 3 1 provided that X 2 + Y 2 + Z 2 = 1 (with equality when ∣ X ∣ = ∣ Y ∣ = ∣ Z ∣ = 3 1 ). The AM/GM Inequality tells us that ∣ X Y Z ∣ 3 2 ≤ 3 1 ( X 2 + Y 2 + Z 2 ) , so that ∣ X Y Z ∣ ≤ 3 3 1 provided that X 2 + Y 2 + Z 2 = 1 (with equality when ∣ X ∣ = ∣ Y ∣ = ∣ Z ∣ = 3 1 ). Thus, subject to the constraint X 2 + Y 2 + Z 2 = 1 , we have ∣ F ∣ ≤ 1 6 7 × 3 3 1 × [ 5 − 3 2 ] = 1 4 4 3 9 1 with equality when ∣ X ∣ = ∣ Y ∣ = ∣ Z ∣ = 3 1 .
Thus we have shown that ( a 7 + b 7 + c 7 + d 7 ) 2 ≤ ( 1 4 4 3 9 1 ) 2 ( a 2 + b 2 + c 2 + d 2 ) 7 = 6 2 2 0 8 8 2 8 1 ( a 2 + b 2 + c 2 + d 2 ) 7 provided that a + b + c + d = 0 , with equality when (for example) a = b = c = 2 3 1 and d = − 2 1 3 , and so we deduce that M = 8 2 8 1 .
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 ( Π ( a + b ) ) ( ( ∑ a 2 + ∑ a b ) 2 + a b c ∑ a )
(To do that,first note that when a = − b ,the expression is equal to 0 ,so it must have a factor ( a + b ) ,and similarly, ( b + c ) and ( c + a ) .)
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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.
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
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The expression is homogeneous, in the sense that if a , b , c , d satisfies the inequality, then so does u a , u b , u c , u d for any u . Thus it is enough to restrict attention to those solutions for which (for example) a 2 + b 2 + c 2 + d 2 takes a particular value, such as 1 .
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 ?
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M = 8 2 8 1 is the smallest value such that the inequality holds for all a , b , c , d with a + b + c + d = 0 . There are certainly some values of a , b , c , d for which a smaller value of M would work, but M = 1 (for example), would not work with a = b = c = 1 , d = − 3
I simply chose a = b = c = 1 / 3 and d = 1 , and proceeded to solve it numerically from there, producting the answer 8 2 9 1 .
... 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 should be − 1 / 3 * 8291 should be 8281
6 2 ′ 2 0 8 M is a maximum of f ( a , b , c , d ) = ( a 2 + b 2 + c 2 + d 2 ) 7 ( a 7 + b 7 + c 7 + d 7 ) 2 , when g ( a , b , c , d ) = a + b + c + d = 0 .
Extrema occur when ∇ f and ∇ g are collinear (where ∇ f is the vector of the partial derivatives of f ).
With S = ( a 7 + b 7 + c 7 + d 7 ) and T = ( a 2 + b 2 + c 2 + d 2 ) , we have ∇ g = ( 1 ; 1 ; 1 ; 1 ) and
∇ f = 1 4 ⋅ S ⋅ T 6 ( T ⋅ a 6 − S ⋅ a ; T ⋅ b 6 − S ⋅ b ; T ⋅ c 6 − S ⋅ c ; T ⋅ d 6 − S ⋅ d ) .
We have collinearity:
when S = 0 (which gives M = 0 for equality, that is clearly a local minimum and not a maximum);
when T = 0 (which gives a = b = c = d = 0 and then M can be anything for equality);
when T ⋅ x 6 − S ⋅ x = T ⋅ y 6 − S ⋅ y , for x and y any of a , b , c or d .
T ⋅ x 6 − S ⋅ x = T ⋅ y 6 − S ⋅ y ⇔ x = y or x − y x 6 − y 6 = T S .
If x − y x 6 − y 6 = T S and x − z x 6 − z 6 = T S , then z = y (because h ( y ) = x − y x 6 − y 6 is always increasing, therefore bijective after checking what occurs at y = x ).
Therefore, a , b , c and d must be the same value x , or be of two distinct values x and y .
If all are the same values, then the sum being 0 implies that all are 0 and that T =0 (case already seen).
If values are x , x , y , y , then the sum being 0 implies that x = − y and therefore S = 0 (case already seen).
If values are x , x , x , y , then the sum being 0 implies that y = − 3 x .
In this extreme case, we have: 6 2 ′ 2 0 8 M = f ( x , x , x , − 3 x ) = ( 3 + ( − 3 ) 2 ) 7 ⋅ x 1 4 ( 3 − 3 7 ) 2 ⋅ x 1 4 = 1 2 7 2 ′ 1 8 4 2
That is, M = 6 2 ′ 2 0 8 ⋅ 1 2 7 2 ′ 1 8 4 2 = 2 8 ⋅ 3 5 ⋅ 2 1 4 ⋅ 3 7 2 6 ⋅ 3 2 ⋅ 7 2 ⋅ 1 3 2 = 7 2 ⋅ 1 3 2 = 8 ′ 2 8 1
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.
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I find f ( 1 , − 2 1 , − 2 1 , 0 ) = 0 . 0 5 6 7 1 3 and M = 3 , 5 2 8 . Am I wrong?
Not a proof, but a fruitful thought. We want to maximize the fraction f = ( a 2 + . . ) 7 ( a 7 + . . ) 2 and we observe that
In our case we get M = 6 2 2 0 8 f m a x = 6 2 2 0 8 × ( 3 2 + 3 ) 7 ( 3 7 − 3 ) 2 = 8 2 8 1
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.
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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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 , we can let d = − a − b − c , so that 6 2 2 0 8 ( 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 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 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 Notice that f ( a , b , c ) is symmetric, meaning that f ( a , b , c ) = f ( b , a , c ) = f ( c , b , a ) = ⋯ for all permutations of ( a , b , c ) . If we consider optimizing 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 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 ⋅ 3 7 ⋅ ( 1 − 3 6 ) 2 ⋅ a 1 4 = 1 2 7 ⋅ M ⋅ a 1 4 = 2 1 4 ⋅ 3 7 ⋅ M ⋅ a 1 4 M = 2 1 4 ⋅ 3 7 2 8 ⋅ 3 7 ⋅ ( 3 6 − 1 ) 2 = 2 6 ( ( 3 3 − 1 ) ( 3 3 + 1 ) ) 2 = 1 3 2 ⋅ 7 2 = 8 2 8 1