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Given a,b,c>0 and a+b+c=3

Prove that :[tex] (a+b^{2})(b+c^{2})(c+a^{2}) \leq 13[/tex]

Prove that :[tex] (a+b^{2})(b+c^{2})(c+a^{2}) \leq 13[/tex]

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- Thread starter phucghe
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- #1

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Given a,b,c>0 and a+b+c=3

Prove that :[tex] (a+b^{2})(b+c^{2})(c+a^{2}) \leq 13[/tex]

Prove that :[tex] (a+b^{2})(b+c^{2})(c+a^{2}) \leq 13[/tex]

- #2

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what's your initial step ?

- #3

uart

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Interesting equation, does anyone have any idea on this one? I've got a hunch that the function's maxima occurs on one of the boundaries (eg a=0) but I'm not sure how to prove it.

BTW. I don't believe that function ever attains 13 on the given domain, it could just as well be written strictly less than 13.

BTW. I don't believe that function ever attains 13 on the given domain, it could just as well be written strictly less than 13.

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matt grime

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arildno

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Let F(a,b,c) equal the left-hand side of your inequality.

It is trivial to show that none of its derivatives equals zero in the region a,b,c>0.

Thus, there are no interior maxima there, so whatever global maximum you'll have will lie somewhere on the boundary.

- #6

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matt grime, the procedure you sketch does not seem to be that easy. The equations for finding stationary points in the interior are quite horrible. Even finding the maxima on the boundary is something I'd rather not do by hand. I wonder if there is not a nicer way to prove the inequality.

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matt grime

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They are? What do you think they should be then?

- #8

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They are? What do you think they should be then?

[tex] (b+c^2)(c+a^2) + 2a(a+b^2)(b+c^2) + \lambda = 0 [/tex]

and the two cyclic permutations of this equation, plus the constraint

[tex] a+b+c=3 [/tex]

where lambda is a Lagrange multiplier.

- #9

matt grime

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Why invoke lagrange multipliers to show that some function doesn't have a local maximum?

- #10

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Why invoke lagrange multipliers to show that some function doesn't have a local maximum?

Because there is a constraint. Even if the function does not have a maximum in the region { a,b,c > 0 }, it may have a maximum in { a,b,c > 0 and a+b+c = 3 }. Of course you can get rid of the constraint by eliminating one of the variables (say a), and instead look for extrema of

[tex] (3-b-c+b^2) (b+c^2) (c+(3-b-c)^2). [/tex]

That looks even more horrible.

- #11

matt grime

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I'd invoke multipliers if I wanted to *find* the maximum.

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- #13

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Edit: this doesn't work. let a = 2.999 b = .0005, c = .0005

whoops that doesnt work sorry

whoops that doesnt work sorry

- #14

uart

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Hi Jitse Niesen, I know what you're saying, I was thinking along the same lines.

It's easy to show that the unconstrained function has no stationary points, but when you add the constraint (a+b+c=3) then it’s a lot more messy.

If you substitute c=3-a-b to make it a function of two variables you get at least two stationary points. One is exactly at (a,b)=(1,1) and there's another somewhere near (a,b)=(0.76,0.74). The first one (1,1) turns out to be a local minima and I'm pretty sure that the second one is a saddle point.

They were pretty messy derivatives and I used maple to help me with those results. The point I was making before was just that if there was some other easy argument (say some sort of symmetry argument or whatever) to show that the maximum occurred when one of the variables was zero then the whole thing could be reduced pretty easily to a one dimensional problem.

BTW. With c=0 and b=3-a then resulting 1d problem gives the following cubic:

[tex]5\,a^3 -13\,a^2 + 15\,a - 9 = 0[/tex].

Numerically this has the (approx) solution**a=1.369**, (and b=1.631, c=0) which gives f_max approx **12.765**.

It's easy to show that the unconstrained function has no stationary points, but when you add the constraint (a+b+c=3) then it’s a lot more messy.

If you substitute c=3-a-b to make it a function of two variables you get at least two stationary points. One is exactly at (a,b)=(1,1) and there's another somewhere near (a,b)=(0.76,0.74). The first one (1,1) turns out to be a local minima and I'm pretty sure that the second one is a saddle point.

They were pretty messy derivatives and I used maple to help me with those results. The point I was making before was just that if there was some other easy argument (say some sort of symmetry argument or whatever) to show that the maximum occurred when one of the variables was zero then the whole thing could be reduced pretty easily to a one dimensional problem.

BTW. With c=0 and b=3-a then resulting 1d problem gives the following cubic:

[tex]5\,a^3 -13\,a^2 + 15\,a - 9 = 0[/tex].

Numerically this has the (approx) solution

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- #15

arildno

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What I said was wrong. Sorry about that.

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