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Title: inequality: a^4 + b^4 + c^4 Post by william wu on Aug 23rd, 2003, 10:00pm Prove the following inequality: [forall]a,b,c[in][bbr] : a4 + b4 + c4 [ge] abc(a + b + c) |
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Title: Re: inequality: a^4 + b^4 + c^4 Post by towr on Aug 24th, 2003, 10:51am partial (maybe [e]not[/e]) [hide] 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][/hide] |
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Title: Re: inequality: a^4 + b^4 + c^4 Post by NickH on Aug 24th, 2003, 4:37pm This follows immediately from the (PDF) Rearrangement Inequality (http://matholymp.com/TUTORIALS/Rear.pdf). |
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Title: Re: inequality: a^4 + b^4 + c^4 Post by towr on Aug 25th, 2003, 1:55am 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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Title: Re: inequality: a^4 + b^4 + c^4 Post by TenaliRaman on Aug 25th, 2003, 9:06am 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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Title: Re: inequality: a^4 + b^4 + c^4 Post by NickH on Aug 25th, 2003, 11:14am 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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Title: Re: inequality: a^4 + b^4 + c^4 Post by TenaliRaman on Aug 25th, 2003, 12:13pm 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+c2ab (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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Title: Re: inequality: a^4 + b^4 + c^4 Post by towr on Aug 25th, 2003, 12:45pm try spaces.. Sometimes a tag get's broken if the spaceless line is too long (dunno why). |
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Title: Re: inequality: a^4 + b^4 + c^4 Post by NickH on Aug 25th, 2003, 12:55pm Quote:
Yes, TenaliRaman, that's what I'm saying. |
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Title: Re: inequality: a^4 + b^4 + c^4 Post by SWF on Aug 26th, 2003, 6:38pm 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: [hide]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.[/hide] Step 2: [hide]This leaves a function that must be shown to be [ge]0: H(a,x,y)=5a2(x2-xy+y2)+a(4x3-x2y-xy2 +4y3)+x4+y4[/hide] Step 3: [hide]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.[/hide] Step 4: [hide]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.[/hide] Step 5:[hide]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):[/hide] Step 6:[hide]4AC-B2= (2x3-3x2y-3xy2+2y3)2 + 22(x2y-xy2)2 + 8x2y2(x2+y2) [ge] 0[/hide] |
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Title: Re: inequality: a^4 + b^4 + c^4 Post by TenaliRaman on Aug 27th, 2003, 11:44am very very neat !! :o |
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