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I am not sure if this question is appropriate here but would appreciate any help. Let $n>2$ and $x_1,\cdots,x_n$ be real numbers. What is the infimum of:

$$A = \sum_{i=1}^n \frac{1}{(1+x_i)^2} $$

subject to the constraint $\prod_{i=1}^nx_i=1$?

Similar question for:

$$B = \sum_{i=1}^n \frac{1}{(1-x_i)^2} $$

Thanks in advance!

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Do you have a positivity condition on the x? These kind of problems often require in addition that x_i > 0. I'm not shure it makes a difference in this case 'though. –  Michael Ulm Aug 9 '10 at 6:41
    
@Michael: I don't. But if that helps, please post your solution. –  curious Aug 9 '10 at 6:45
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1 Answer

up vote 3 down vote accepted

I assume you intend for all the $x_i$ to be positive. The infimum is $\min(n/4, 1)$.

This is a good example of how to use convexity to minimize an expression.

Make the change of variables $x_i = e^{t_i}$, so $\sum t_i=0$. Define $f(t) = 1/(1+e^t)^2$. We want to minimize $\sum f(t_i)$.

Now, $f(t)$ is convex for $t> - \log 2$ and concave for $t < - \log 2$. This is an easy calculation with the double derivative of $f$. (You might find this graph helpful.) So, if all of the $t_i$ lie in $[- \log 2, \infty)$, then Jensen's inequality shows that the minimum occurs when all the $t_i$ are $0$. In that case, we get $n/4$.

We claim that we can restrict to the case that at most one of the $t_i$ lies in $(- \infty, - \log 2)$. Proof: Consider any solution $(t_i)$, with $t_1$, $t_2$, ..., $t_k < - \log 2$. Replace these values by $k-1$ copies of $- \log 2$ and $t_1+t_2+\cdots+t_k + (k-1) \log 2$. Since $f$ is concave in this range, we have decreased the sum.

So, we may assume that there is one $t_i$, call it $u$, in $(- \infty, - \log 2)$, and all the others are in $[- \log 2, \infty)$. For a fixed value of $u$, by convexity, we will obtain the minimum when all the other values are equal to each other, say $v$. So $u+(n-1)v=0$. Our sum is $$g(v) :=(1+e^{-(n-1)v})^{-2} + (n-1) (1+e^v)^{-2}.$$ We want to minimize this quantity, subject to the condition that $v \geq -\log 2$.

As $v \to \infty$, the function $g$ goes to $1$. We'd like to know whether there is any point in $[\log 2, \infty)$ where $g$ dips below $1$. Such a point must be a local minimum. Here are some plots for $n$ equal to 3 and equal to 10. Numerically, it looks like there is a local minimum at $0$, where $g$ is $n/4$ and a local maximum at some positive value. If we can confirm this, the infimum is $\min(1, n/4)$.

Let's put the cherry on this sundae and prove that there is no other local minimum. The derivative $g'$ is $$\frac{-2(n-1)e^{-(n-1)v}}{(1+e^{-(n-1)v})^3} + \frac{(n-1) e^v}{(1+e^v)^3}$$ Set this equal to zero and clear out denominators to get $$(1+e^v)^3 e^{2 (n-1)v} = (1+e^{(n-1)v})^3 e^v.$$

Set $w=e^v$. For $n \geq 4$, this is a polynomial in $w$ with two sign changes so, by Descartes' rule of signs, it has at most two positive roots. There also turn out to be two real roots for $n=2$ and $3$, by computer check. These are the local minimum we have already noticed at $v=0$ ($w=1$), and the local maximum nearby. So there are no other local minima, and we are done.

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It looks like all my links to Wolfram alpha have been stripped out for some reason. (Markup bug?) Anyway, if you want to see plots of $f$ and $g$, it's not hard to generate them for yourself. –  David Speyer Aug 9 '10 at 14:21
    
Thanks, KennyTM! –  David Speyer Aug 9 '10 at 14:37
    
David, thanks a lot for the great answer! I can't upvote yet, but will do so when I have enough reputation. –  curious Aug 9 '10 at 18:15
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