Infimum of some amounts

I am not exactly sure if this inquiry is ideal below yet would certainly value any kind of aid. Allow $n>2$ and also $x_1,\cdots,x_n$ be actual numbers. What is the infimum of:

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

based on the restraint $\prod_{i=1}^nx_i=1$?

Comparable inquiry for:

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

Many thanks beforehand!

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2019-05-06 22:37:31
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Answers: 1

I think you plan for all the $x_i$ to be favorable. The infimum is $\min(n/4, 1)$.

This is an example of just how to make use of convexity to decrease an expression.

Make the adjustment of variables $x_i = e^{t_i}$, so $\sum t_i=0$. Specify $f(t) = 1/(1+e^t)^2$. We intend to decrease $\sum f(t_i)$.

Currently, $f(t)$ is convex for $t> - \log 2$ and also concave for $t < - \log 2$. This is a very easy estimation with the double by-product of $f$. (You could locate this graph handy.) So, if every one of the $t_i$ hinge on $[- \log 2, \infty)$, after that Jensen's inequality reveals that the minimum takes place when all the $t_i$ are $0$. Because instance, we get $n/4$.

We assert that we can limit to the instance that at the majority of among the $t_i$ hinges on $(- \infty, - \log 2)$. Evidence : Consider any kind of remedy $(t_i)$, with $t_1$, $t_2$, ..., $t_k < - \log 2$. Change these values by $k-1$ duplicates of $- \log 2$ and also $t_1+t_2+\cdots+t_k + (k-1) \log 2$. Given that $f$ is concave in this array, we have actually lowered the amount.

So, we might think that there is one $t_i$, call it $u$, in $(- \infty, - \log 2)$, and also all the others remain in $[- \log 2, \infty)$. For a dealt with value of $u$, by convexity, we will certainly get the minimum when all the various other values amount to each various other, claim $v$. So $u+(n-1)v=0$. Our amount is $$g(v) :=(1+e^{-(n-1)v})^{-2} + (n-1) (1+e^v)^{-2}.$$ We intend to decrease this amount, based on the problem that $v \geq -\log 2$.

As $v \to \infty$, the function $g$ mosts likely to $1$. We would certainly such as to recognize whether there is any kind of factor in $[\log 2, \infty)$ where $g$ dips listed below $1$. Such a factor has to be a neighborhood minimum. Below are some stories for $n$ equal to 3 and also equal to 10. Numerically, it resembles there is a neighborhood minimum at $0$, where $g$ is $n/4$ and also a neighborhood maximum at some favorable value. If we can validate this, the infimum is $\min(1, n/4)$.

Allow's place the cherry on this sundae and also confirm that there is nothing else neighborhood minimum. The acquired $g'$ 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 equivalent to absolutely no and also clear out 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 2 indicator adjustments so, by Descartes' rule of signs, it contends the majority of 2 favorable origins. There additionally become 2 actual favor $n=2$ and also $3$, by computer system check. These are the neighborhood minimum we have actually currently seen at $v=0$ ($w=1$), and also the neighborhood maximum close by. So there are nothing else neighborhood minima, and also we are done.

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2019-05-09 10:57:19
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