Trying to prove inequality with Lagrange multipliers

[xman]

Show that if we have N positive numbers
\left[ p_{i}\right]_{i=1}^{N}
such that
\sum_{i} p_{i} =1
then for any N numbers
\left\{x_{i}\right\}_{i=1}^{N}
we have the inequality
\prod_{i=1}^{N} x_{i}^{2 p_{i}} \leq \sum_{i=1}^{N} p_{i}x_{i}^{2}

So I am thinking to show the inequality is true using Lagrange multipliers first take the set
W = \sum_{i} p_{i}x_{i}^{2}
and we want to minimize above subject to constraint
S = \prod_{i} x_{i}^{2p_{i}}
so we form the function
f^{\star} = f + \lambda g \Rightarrow f^{\star} =\sum_{i} p_{i}x_{i}^{2}+\lambda \left(S-\prod_{i} x_{i}^{2p_{i}}\right)
So I think everything so far is ok...my question is how do you differentiate an infinite series and an infinite product. Also in this case is the Lagrange multiplier a single value \lambda or is there one multiplier for each value of i , that is; do I need a \lambda_{i} Any direction or input is greatly appreciated.

 

[StatusX]

I'm a little confused by some of your wording, but I think what you're trying to do is show that, for a fixed set of p_i, and a fixed value of S, W is always greater than or equal to S, whatever the x_i that give this S may be. If you show this is true for any S, then for a given set of x_i, these give rise to some S, and then you know the corresponding W for these x_i must be greater than or equal to S. That's an interesting approach. I'm not sure if it'll work, but it's worth a try.

You'll only need one \lambda, because there's only one constraint equation. But you need to differentiate with respect to each x_i. Namely, you have:

\frac{\partial}{\partial x_i} \left( f + \lambda g \right) = 0

for all i from 1 to N.

 

[xman]

Thanks for replying StatusX,
Sorry if I wasn't a little more clear, but that's exactly what I'm trying to show. I thought this would be a fun problem, and I am not really familiar with Lagrange multipliers, so I thought I would try proving it, this way. So, only one common multiplier, great. Now is this correct
\frac{\partial}{\partial x_{j}} \left(p_{i} x_{i}^{2}\right)= 2p_{i}x_{i} \frac{\partial x_{i}}{\partial x_{j}} \delta_{ij}
where \delta_{ij} is the Kronecker delta of course. Now for the infinite product I'm not sure.

 

[StatusX]

First off, you want:

\frac{\partial x_i}{\partial x_j}= \delta_{ij}

since the x_i are independent. And, just for the record, neither the product nor the sum are infinite, they both range from 1 to N.

To differentiate the product, just write it out. All the terms that don't involve x_i will be constants when you differentiate with respect to x_i. You will end up with the original product times some prefactor.

 

[xman]

First off, you want:

\frac{\partial x_i}{\partial x_j}= \delta_{ij}

since the x_i are independent. And, just for the record, neither the product nor the sum are infinite, they both range from 1 to N.

To differentiate the product, just write it out. All the terms that don't involve x_i will be constants when you differentiate with respect to x_i. You will end up with the original product times some prefactor.

Right, I meant to write
\frac{\partial x_{i}}{\partial x_{j}}=\delta_{ij}
Sorry I keep saying "infinite " for product and sum. Ok, so is this correct
2 \sum_{i=1}^{N}p_{i} x_{i}+ \lambda \left(S-2 \prod_{i=1}^{N} p_{i}x_{i}^{2p_{i}-1} \right) =0
Does this seems reasonable?

 

[StatusX]

No, you only want to differentiate with respect to one x_i at a time. You'll get N different equations.

 

[xman]

No, you only want to differentiate with respect to one x_i at a time. You'll get N different equations.

Sorry, I'm a little uncomfortable with these differentiation rules. Here it goes so I should get
2 p_{j} x_{j}+ \lambda \left( \frac{\partial S}{\partial x_{j}}-2 p_{j} x_{j}^{2p_{j}-1} \left( \prod_{i <j} x_{i}^{2p_{i}}\right)\right)=0
where 1<j<i<N

 

[StatusX]

Closer. If you replaced i<j by i≠j, you'd just about have it, although there's a more convenient way to write it (in terms of S). And remember, S is a constant.

 

[xman]

Great so it would be something like\ldots

2 p_{j} x_{j}+ \lambda \left( -2 p_{j} x_{j}^{2p_{j}-1} \left( \prod_{i \neq j} x_{i}^{2p_{i}}\right)\right)=0

From here we solve for \lambda with the requirement that we want to minimize, right?

 

[StatusX]

Well, if you rewrite the last term in terms of S, like I suggested, you'll see that the equation is the same for every x_i. What does this tell you? (You don't need to know lambda)

 

[xman]

Oh snap are you saying something along the lines of
<br /> \frac{\partial f^{\star}}{\partial x_{1}} = 2 p_{1} x_{1}<br /> - 2\lambda p_{1}x_{1}^{2p_{1}-1}\prod_{i=2}^{N} x_{i}^{2p_{i}}=0 \cr<br /> \ldots \\<br /> \frac{\partial f^{\star}}{\partial x_{N}} = 2p_{N} x_{N}<br /> -2\lambda p_{N}x_{N}^{2p_{N}-1} \prod_{i=1}^{N-1}<br /> x_{i}^{2p_{i}}=0 \\<br />
So
<br /> \Rightarrow<br /> \frac{\partial f^{\star}}{\partial x_{1}} = p_{1} x_{1}<br /> -\lambda p_{1}x_{1}^{-1}S=0<br /> \ldots<br /> \frac{\partial f^{\star}}{\partial x_{N}} = p_{N} x_{N}<br /> -\lambda p_{N}x_{N}^{-1}S<br /> \\<br /> \Rightarrow<br /> \sum_{i} p_{i} x_{i}^{2} = \lambda \left(\sum_{i}p_{i}\right)S<br />
Thus
<br /> \Rightarrow<br /> S^{-1} \sum_{i} p_{i} x_{i}^{2} = \lambda<br />

 

[StatusX]

I was with you up till the last line. Try writing an expression for each x_i in terms of only lambda and S. The exact equation isn't important, what is important is that it is the same for all x_i, which means... (the key point).

 

[xman]

So we have
p_{1} x_{1}=\lambda \frac{p_{1}S}{x_{1}} \ldots p_{N}x_{N}= \lambda<br /> \frac{p_{N} S}{x_{N}}
which simplifies to
x_{1} = \lambda \frac{S}{x_{1}} \ldots<br /> x_{N} = \lambda \frac{S}{x_{N}} \Rightarrow x_{1}^{2}=\lambda<br /> S\ldots x_{N}^{2}=\lambda S
Right? I guess I'm obviously missing the key point here, the only thing that comes to mind now is
\lambda = \frac{x_{1}^{2}}{S} = \ldots =\frac{x_{N}^{2}}{S}
So
x_{1}^{2}=\ldots = x_{N}^{2}
So
<br /> \prod_{i=1}^{N} \left(x_{i}^{2}\right)^{p_{i}} \Rightarrow<br /> \left(x_{N}^{2}\right)^{\sum_{i}p_{i}} = x_{N}^{2}
since we are given that
\sum_{i} p_{i} =1
Am I heading down the wrong path again?

 

[StatusX]

Right, you needed to show they were all equal. What this means is that W is at an extreme value (max or min) when all of the x_i are equal. Now just calculate what S are W are in this case, and verify the inequality. You then need to verify this is actually the minimum of W.

 

[xman]

Right, you needed to show they were all equal. What this means is that W is at an extreme value (max or min) when all of the x_i are equal. Now just calculate what S are W are in this case, and verify the inequality. You then need to verify this is actually the minimum of W.

Awesome, so I want to show that W is a minimum here. Hence,
W(x^{c}) \leq W(x^{c}+\epsilon) \Rightarrow<br /> x^{2} \left(\sum_{i=1}^{N}p_{i}\right) \leq \left(x+\epsilon\right)^{2} \left(\sum_{i=1}^{N} p_{i}\right)
Yielding
0\leq \epsilon \left(2x+\epsilon\right)
which shows us that for points to the left we have a negative slope and for points to the right we have a positive slope since \epsilon &gt;0 the sign of the equation is dominated by the x-which indeed is a minimum. Now for the inequality it suffices to show
\prod_{i=1}^{N} x_{i}^{2p_{i}} \leq \sum_{i=1}^{N} p_{i}x_{i}^{2} \mid_{x=x^{c}}
which immediately reduces to equality as
x^{2}=x^{2}
per my previous post. Finally, we conclude since W is a minimum, and we have equality at the critical point x^{c} = x_{1}=\cdots=x_{n} then this is indeed an upper bound for our S and therefore the inequality is true.

Is there any points I missed in the wrap up here?