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

  • Thread starter phucghe
  • Start date
In summary, the conversation discusses a proof for the inequality (a+b^{2})(b+c^{2})(c+a^{2}) \leq 13 given a,b,c>0 and a+b+c=3. The initial step is discussed and it is suggested that the function's maximum may occur on one of the boundaries. It is also mentioned that there may not be a maximum of 13 on the given domain. The use of Lagrange multipliers is suggested to show that there are no local maxima in the interior of the region. However, it is also mentioned that this approach may be messy and another argument, such as a symmetry argument, could be used to reduce it to a one-dimensional problem. It is also mentioned
  • #1
phucghe
4
0
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]
 
Mathematics news on Phys.org
  • #2
what's your initial step ?
 
  • #3
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.
 
Last edited:
  • #4
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
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
They are? What do you think they should be then?
 
  • #8
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.
 
  • #9
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.
 
Last edited:
  • #15
What I said was wrong. Sorry about that.
 

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

What is the meaning of the inequality in this problem?

The inequality, $(a+b^2)(b+c^2)(c+a^2) \leq 13$, means that the product of three expressions, $(a+b^2)$, $(b+c^2)$, and $(c+a^2)$, is less than or equal to 13.

What are the variables and their constraints in this problem?

The variables in this problem are a, b, and c. The only constraint is that they are real numbers.

How can I prove this inequality?

There are various methods to prove an inequality, such as algebraic manipulation, induction, or using known theorems. The specific method for this problem would depend on the context and purpose of the proof.

What are the possible values for the expressions $(a+b^2)$, $(b+c^2)$, and $(c+a^2)$?

Since a, b, and c are real numbers, the expressions $(a+b^2)$, $(b+c^2)$, and $(c+a^2)$ can take on any real value. However, in order for the inequality to hold, their product must be less than or equal to 13.

How can this inequality be applied in real-life situations?

Inequalities are often used in real-life situations to model relationships between quantities. This particular inequality may be useful in areas such as economics, statistics, or physics, where the product of multiple expressions needs to be limited to a certain value.

Similar threads

MHB Prove $a^2+b^2-c^2>6(c-a)(c-b) for $a^3+b^3=c^3$
Replies
1
Views
1K
MHB -c5.LCM and Prime Factorization of A,B,C
Replies
3
Views
819
MHB Basic Inequality Prove: $A\leq B\wedge B\leq A \Rightarrow A=B$
Replies
2
Views
907
B How to pick some numbers out of 13 integers, by a 4 digits code
Replies
4
Views
1K
MHB Prove Triangle Inequality: $\sqrt{2}\sin A-2\sin B+\sin C=0$
Replies
2
Views
1K
MHB Proving Equivalence of 0A=0, (-A)B=-AB, (-A)(-B)=AB
Replies
1
Views
842
MHB Triangle Equality Proof: $\frac{a}{b}=1+\sqrt{2\left(\frac{c}{b}\right)^2-1}$
Replies
1
Views
1K
MHB Inequality c≤√[(x−a^2+(y−b)^2+(z−c)^2]+√(x2+y2+z2)
Replies
5
Views
1K
B When are two vectors collinear and coplanar?
Replies
4
Views
356
MHB Prove $b^2-c^2\leq2a^2$ in $\triangle ABC$
Replies
1
Views
959

Hot Threads

Recent Insights

Back
Top