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        riddles >> putnam exam (pure math) >> inequality: a^4 + b^4 + c^4
        
(Message started by: william wu on Aug 23^rd, 2003, 10:00pm)

Title: inequality: a^4 + b^4 + c^4
Post by william wu on Aug 23^rd, 2003, 10:00pm

Prove the following inequality:

[forall]a,b,c[in][bbr] : a⁴ + b⁴ + c⁴ [ge] abc(a + b + c)

Title: Re: inequality: a^4 + b^4 + c^4
Post by towr on Aug 24^th, 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]

Title: Re: inequality: a^4 + b^4 + c^4
Post by NickH on Aug 24^th, 2003, 4:37pm

This follows immediately from the (PDF) Rearrangement Inequality (http://matholymp.com/TUTORIALS/Rear.pdf).

Title: Re: inequality: a^4 + b^4 + c^4
Post by towr on Aug 25^th, 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

Title: Re: inequality: a^4 + b^4 + c^4
Post by TenaliRaman on Aug 25^th, 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.

Title: Re: inequality: a^4 + b^4 + c^4
Post by NickH on Aug 25^th, 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.)

Title: Re: inequality: a^4 + b^4 + c^4
Post by TenaliRaman on Aug 25^th, 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

Title: Re: inequality: a^4 + b^4 + c^4
Post by towr on Aug 25^th, 2003, 12:45pm

try spaces..
Sometimes a tag get's broken if the spaceless line is too long (dunno why).

Title: Re: inequality: a^4 + b^4 + c^4
Post by NickH on Aug 25^th, 2003, 12:55pm

Quote:

is this equivalent to what you r saying NickH

Yes, TenaliRaman, that's what I'm saying.

Title: Re: inequality: a^4 + b^4 + c^4
Post by SWF on Aug 26^th, 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]

Title: Re: inequality: a^4 + b^4 + c^4
Post by TenaliRaman on Aug 27^th, 2003, 11:44am

very very neat !! :o