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Topic: inequality: a^4 + b^4 + c^4 (Read 9922 times) |
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towr
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Re: inequality: a^4 + b^4 + c^4
« Reply #1 on: Aug 24th, 2003, 10:51am » |
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partial (maybe [e]not[/e])
a4 + b4 + c4 >= a4 + b4 + c4 - 2(a2 + b2 + c2 ) = (a+b+c)(a-b+c)(a+b-c)(a-b-c) >= abc(a+b+c)
so it's left to be proven that
(a-b+c)(a+b-c)(a-b-c) >= abc
[edit]false when a=b=c=1, so nevermind this approach[/edit]
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| « Last Edit: Aug 25th, 2003, 2:12am by towr » |
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towr
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Re: inequality: a^4 + b^4 + c^4
« Reply #3 on: Aug 25th, 2003, 1:55am » |
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How?
I can see
a4 + b4 + c4 >= a3b + b3c + a c3
follows from it but
a3b + b3c + c3a >= a2b c + a b2c + a b c2
doesn't hold true, try f.i. a=-1, b=1, c=0
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TenaliRaman
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Re: inequality: a^4 + b^4 + c^4
« Reply #4 on: Aug 25th, 2003, 9:06am » |
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I did start with the rearrangement for this one but then didn't get much anywhere.I even tried some variations like
a4+b4+c4>=a2b2 + b2c2 + c2a2
then tried to use b<c and a<c props to get
a4+b4+c4>=a2b2 + b2ac + a2bc
still one term left to modify 
then i tried to use AM - GM on the right hand side to convert it to 3*(abc)3/2 .. i thought this might simplify things .... and i am still working on it.
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NickH
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Re: inequality: a^4 + b^4 + c^4
« Reply #5 on: Aug 25th, 2003, 11:14am » |
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Consider the two sets of ordered triplets:
{(a,b,c), (a,b,c), (a,b,c), (a,b,c)}
{(a,b,c), (a,b,c), (b,c,a), (c,a,b)}
The "dot product" (what's the proper term here?) of each set of triplets is, respectively:
a4 + b4 + c4
a2bc + b2ca + c2ab
Now the result follows from the rearrangement inequality. (Or a slight extension thereof, that makes use of more than two ordered n-tuplets.)
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| « Last Edit: Aug 25th, 2003, 11:19am by NickH » |
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TenaliRaman
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Re: inequality: a^4 + b^4 + c^4
« Reply #6 on: Aug 25th, 2003, 12:13pm » |
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Great Idea Nick!!!!
is this equivalent to what you r saying NickH,
a<b<c rearrangement (a,b,c)
a<b<c rearrangement (a,b,c)
a<b<c rearrangement (b,c,a)
a<b<c rearrangement (c,a,b)
By rearrangement inequality,
a4+b4+c4>=a2bc+b2ca+c[s up]2[/sup]ab
(surprisingly small for a proof!!  )
P.S something seems to be a prob with the last superscript ... it doesn't work out
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| « Last Edit: Aug 25th, 2003, 12:18pm by TenaliRaman » |
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towr
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Re: inequality: a^4 + b^4 + c^4
« Reply #7 on: Aug 25th, 2003, 12:45pm » |
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try spaces..
Sometimes a tag get's broken if the spaceless line is too long (dunno why).
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| « Last Edit: Aug 25th, 2003, 12:45pm by towr » |
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NickH
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Re: inequality: a^4 + b^4 + c^4
« Reply #8 on: Aug 25th, 2003, 12:55pm » |
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Quote:| is this equivalent to what you r saying NickH |
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Yes, TenaliRaman, that's what I'm saying.
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SWF
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Re: inequality: a^4 + b^4 + c^4
« Reply #9 on: Aug 26th, 2003, 6:38pm » |
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With some ingenuity, this problem can also be done with algebra. I will hide the various steps in case somebody wants to work it out with some hints. (Edit.. I see the new types of tags are not hidden, consider those an additional clue).
Step 1: Let x=b-a and y=c-a and move the right side of the inequality to the left to give an expression that must be shown to be [ge]0. Replace b by a+x, c by a+y and expand the terms.
Step 2: This leaves a function that must be shown to be [ge]0:
H(a,x,y)=5a2(x2-xy+y2)+a(4x3-x2y-xy 2 +4y3)+x4+y4
Step 3: Regardless of the values of x and y this is clearly [ge]0 when a=0, and only equals zero when x=y=0. Show that by varying the variable a, with a fixed x and y, H(a,x,y) can never be negative.
Step 4: Use quadratic formula. If 4AC-B2>0, there is no real value of a that makes H(a,x,y) negative. Because either x=y=0 and H(a,x,y)=0, or H(0,x,y)[ge]0, 4AC-B[sup2]>0 and the continuous function, H, never equals zero, so it can't cross over from positive to negative.
Step 5:4AC-B2=4x6 - 12x5y + 27x4y2 - 18x3y3 + 27x2y4 - 12xy5 + 4y6
This is obviously [ge]0 if you reorganize in a clever way (the fun part of this problem):
Step 6:4AC-B2= (2x3-3x2y-3xy2+2y3)2 + 22(x2y-xy2)2 + 8x2y2(x2+y2) [ge] 0
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| « Last Edit: Aug 27th, 2003, 4:07pm by SWF » |
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TenaliRaman
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Re: inequality: a^4 + b^4 + c^4
« Reply #10 on: Aug 27th, 2003, 11:44am » |
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very very neat !! 
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