An easy ineq

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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]
 

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  • #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.
 
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  • #4
matt grime
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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
 
  • #5
arildno
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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.
 
  • #6
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.
 
  • #7
matt grime
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They are? What do you think they should be then?
 
  • #8
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
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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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
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Edit: this doesn't work. let a = 2.999 b = .0005, c = .0005

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.
 
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  • #15
arildno
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What I said was wrong. Sorry about that.
 

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