Prove: $(a+b^2)(b+c^2)(c+a^2) \leq 13$

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Discussion Overview

The discussion revolves around proving the inequality \((a+b^2)(b+c^2)(c+a^2) \leq 13\) under the constraints that \(a, b, c > 0\) and \(a+b+c=3\). Participants explore various approaches to tackle this problem, including considerations of maxima and the implications of constraints.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants suggest that the function's maxima may occur on the boundaries of the defined region, such as when one variable approaches zero.
  • There is a proposal that the function does not attain the value of 13 within the given constraints, potentially being strictly less than 13.
  • One participant mentions that there are no local interior maxima, implying that any global maximum must be found on the boundary.
  • Another participant points out the complexity of finding stationary points in the interior due to the constraint \(a+b+c=3\), suggesting that the process is not straightforward.
  • There is a discussion about using Lagrange multipliers to analyze the problem, with some questioning the necessity of this approach for proving the absence of local maxima.
  • One participant shares findings from substituting variables to reduce the problem's dimensionality, identifying potential stationary points and discussing their nature (local minima, saddle points).
  • Another participant acknowledges an error in their previous reasoning regarding the problem setup.

Areas of Agreement / Disagreement

Participants express differing views on the existence of local maxima and the effectiveness of various mathematical approaches. There is no consensus on a definitive method to prove the inequality or on the nature of the function's extrema.

Contextual Notes

Participants note the complexity introduced by the constraint \(a+b+c=3\) and the challenges in finding stationary points, indicating that the mathematical steps involved may be cumbersome and not easily solvable by hand.

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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]
 
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what's your initial step ?
 
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.
 
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Any global maximum occurs either at an interior local maximum, or on the boundary of the region. I think it is straightforward to show that there are no local interior maxima
 
To OP:
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.
 
arildno, there is also the constraint a+b+c=3 to take into account.

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.
 
They are? What do you think they should be then?
 
matt grime said:
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.
 
Why invoke lagrange multipliers to show that some function doesn't have a local maximum?
 
  • #10
matt grime said:
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
It is still a true (and the point of Lagrangian multipliers) that maxima are either local or on the boundary of the relevant part of the plane a+b+c=3 inside 3-space.

I'd invoke multipliers if I wanted to *find* the maximum.
 
  • #12
I don't understand you. Are you saying that any local maxima of f (the left-hand side of the inequality) restricted to T (the triangular region from the problem) lies either in the interior or the boundary of T? I agree with that, but I don't see how you want to prove that f restricted to T has no local maxima in the interior of T. Could you please explain that, or if you mean something else, explain what you mean?
 
  • #13
Edit: this doesn't work. let a = 2.999 b = .0005, c = .0005

whoops that doesn't work sorry
 
  • #14
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.
 
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  • #15
What I said was wrong. Sorry about that.
 

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