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Let $(a_1,\space a_2,\space \cdots, \space a_n) \in \mathbb R^n_+$ such that $\displaystyle \prod^n_{i=1 }a_i = 1$. Prove that $$\displaystyle \prod^n_{i=1} (1+a_i^2) \le \cfrac {2^n}{n^{2n-2}}\left (\sum^n_{i=1} a_i\right)^{2n-2}$$

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@FOla Yinka:If $4\vert n$,consider a sequence $(a_1,\cdots,a_n)=(1,1,\cdots,-1,-1)$ with $n/2$ 1s and $n/2$ -1s.Then $LHS>0$ and $RHS=0$.In other words,the inequality fails! –  y zhao Oct 29 '12 at 7:28
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I apologize to everyone, $a_1,\cdots ,a_n$ are positive. –  user31280 Oct 29 '12 at 11:31
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While including what one tried and where these tries failed might not be mandatory for not-homework questions, it is definitely recommended. Not doing so (and, in the present case, refusing to do so) is a clear message to (at least some) potential answerers. Are you sure you want to convey this message? –  Did Oct 29 '12 at 11:38
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I tried substituting $a_i=b_i^n$ and homogenizing to get $\prod (\prod b_i^2 + b_i^{2n})/2) \leq (\sum (b_i^n)/n)^{2n-2} \prod b_i^2$, which is then an inequality of symmetric polynomials. It does not seem to follow readily from Muirhead's, and I'm not so versed in more advanced ones. –  Max Nov 11 '12 at 19:08
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This question is hard enough that including failed partial attempts would probably not be helpful. –  Nick Alger Nov 16 '12 at 18:36
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4 Answers

As a partial answer, consider the polynomial: $$P(x)=\prod_{k=1}^{n}(x+a_k)=\sum_{j=0}^{n}x^{n-j} \binom{n}{j} S_j(a_1,\ldots,a_n),$$ where $\binom{n}{j} S_j(a_1,\ldots,a_n)$ is the $j$-th elementary symmetric polynomial in the variables $(a_1,\ldots,a_n)$. By hypothesis we have $S_n=1$. Moreover: $$ (\clubsuit)\quad P(i)P(-i)=\prod_{k=1}^{n}(1+a_k^2)=\left(\sum_{j=0}^{\lfloor n/2 \rfloor}(-1)^j\binom{n}{2j}S_{2j}\right)^2+ \left(\sum_{j=0}^{\lfloor (n-1)/2 \rfloor}(-1)^j\binom{n}{2j+1}S_{2j+1}\right)^2,$$ and by Mac Laurin's inequality we have $$ S_1\geq S_2^{1/2}\geq\ldots\geq S_n^{1/n}=1.$$ Are someone able to derive $$ P(i)P(-i)\leq 2^n\,S_1^{2n-2} $$ from $(\clubsuit)$?

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I have not solved the problem completely, but I have found how to reduce it to an easier-looking problem, in two variables instead of $n$. I use the Lagrange multiplier method. Perhaps someone else can fill the blanks.

Let $\Omega=\lbrace (x_1,x_2, \ldots ,x_n) | x_i >0 \ (1 \leq i\leq n) \rbrace$, and let $M>0$. Consider the following optimization problem on $\Omega$ :

Maximize $f (x_1,x_2, \ldots ,x_n)$ subject to $g(x_1,x_2, \ldots ,x_n)=h(x_1,x_2, \ldots ,x_n)=0$, where

$$ \begin{array}{lcl} f (x_1,x_2, \ldots ,x_n) &=& (1+x_1^2)(1+x_2^2) \ldots (1+x_n^2) \\ g(x_1,x_2, \ldots ,x_n) &=& x_1+x_2+ \ldots +x_n-M \\ h(x_1,x_2, \ldots ,x_n) &=& x_1x_2 \ldots x_n-1 \\ \end{array} $$

The set $K_M=\lbrace (x_1,x_2, \ldots ,x_n) | g(x_1,x_2, \ldots ,x_n)=h(x_1,x_2, \ldots ,x_n)=0\rbrace$ is closed and bounded in the open set $\Omega$. So $f$ attains a minimum on $K_M$ at a point $p$, and we then know that Lagrange multipliers exist, i.e. there are two constants $\lambda$ and $\mu$ such that

$$ \frac{\partial f}{\partial x_k}(p)=\lambda \frac{\partial g}{\partial x_k}(p)+\mu \frac{\partial h}{\partial x_k} \ (1 \leq k \leq n) $$

In other words,

$$ (1) \ 2x_k\prod_{j\neq k}(1+x_j^2)=\lambda+\mu(\prod_{j\neq k}x_j) \ (1 \leq k \leq n) $$

Now (1) is a linear system of $n$ equations in two unknowns $\lambda$ and $\mu$. If $i$ and $j$ are indices such that $x_j \neq x_i$, using the $i$-th and the $j$-th equation we can solve for $\lambda$ and $\mu$ :

$$ (2) \ \lambda=\frac{2(x_i+x_j)}{(x_i^2+1)(x_j^2+1)}\frac{A}{B}, \ \ \ \mu=\frac{x_ix_j(x_ix_j-1)}{(x_i^2+1)(x_j^2+1)}\frac{A}{B} \ (\text{where} \ A=\prod_{j=1}^{n}(1+x_j^2), B= \prod_{j=1}^{n}x_j) $$

Since (2) must hold for all pairs of indices $i,j$ with $x_i \neq x_j$, we deduce that either $\lbrace x_k \rbrace$ has at most two elements, or $n=3$ and $x_1x_2+x_1x_3+x_2x_3=1$.

Now this latter case is impossible since it would imply $$ (x_1-x_2)^2+(x_1-x_3)^2+(x_2-x_3)^2=2((x_1+x_2+x_3)^2-(x_1x_2+x_1x_3+x_2x_3 ))=0 $$ and hence all the $x_i$ would be equal to $1$ (as $x_1x_2x_3=1$),contradicting $x_1x_2+x_1x_3+x_2x_3=1$.

So $\lbrace x_k \rbrace$ consists of at most two elements $a$ and $b$ (with possibly $a=b$). Denoting the number of occurrences by $n_a$ and $n_b$, we thus have $n_a+n_b=n$ and

$$ \begin{array}{l} f(x_1, \ldots ,x_n)=(1+a^2)^{n_a}(1+b^2)^{n_b}, \\ 1=x_1x_2x_3 \ldots x_n=a^{n_a}b^{n_b}, \\ x_1+x_2+ \ldots+x_n=n_aa+n_bb \end{array} $$

It seems that new ideas are needed at this point to finish the solution.

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you swapped $\lambda$ with $\mu$ after differentiating. –  user31280 Nov 16 '12 at 18:46
    
@F'OlaYinka : corrected, thanks. –  Ewan Delanoy Nov 16 '12 at 19:05
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A synthesis of answers proposed by Jack D'Aurizio and Ewan Delanoy may lead to a possible solution to this question.

Step1: If $a_i$ tends to infinity,then

$$\displaystyle \prod^n_{i=1} (1+a_i^2) /\left (\sum^n_{i=1} a_i\right)^{2n-2}$$

bounded above by a constant smaller than $2^n/n^{2n-2}$.

Proof:

Lemma:If $a_i,a_j\geq1$ or $0<a_i,a_j\leq1$,then $(1+a_i^2)(1+a_j^2)\leq2(1+a_i^2a_j^2)$.

It follows from direct calculation.

By our lemma and AM-GM inequality we can get

$$\displaystyle \prod^n_{i=1} (1+a_i^2) /\left (\sum^n_{i=1} a_i\right)^{2n-2}\leq 2^{n-2}(1+A^2)(1+A^{-2})/(mA^{\frac{1}{m}}+(n-m)A^{-\frac{1}{n-m}})^{2n-2}$$

where $A$ is the product over all the $m$ positive real numbers $a_i\geq 1$.

Suppose some $a_i\rightarrow\infty$ and $m<n-1$.Then $A$ tends to infinity and the $RHS$ of the inequality above tends to zero.A much more difficult case is $m=n-1$.

Recall that $(1+a_i^2)<(1+a_i)^2$.Then

$$\displaystyle \prod^n_{i=1} (1+a_i^2) /\left (\sum^n_{i=1} a_i\right)^{2n-2}\leq \displaystyle (\prod^n_{i=1} (1+a_i) /\left (\sum^n_{i=1} a_i\right)^{n-1})^2$$

From Jack D'Aurizio's discussion, $$\prod_{k=1}^{n}(1+a_k)=\sum_{j=0}^{n} \binom{n}{j} S_j$$.

Newton's inequality implies

$$S_j\leq S_1^j$$

and $S_j/S_1^{n-1}=n^{n-1}S_j/\left (\sum^n_{i=1} a_i\right)^{n-1}$ tends to zero when some $a_i$ tend to infinity.

Recall the assumption that $m=n-1$.Then $$nS_{n-1}+S_n=1+\sum_{k=1}^n\frac{1}{a_k}\leq n+\prod_{a_i>=1}{a_i}<n+(\frac{\sum_{i=1}^n a_i}{n-1})^{n-1}$$.

So $$\displaystyle \prod^n_{i=1} (1+a_i^2) /\left (\sum^n_{i=1} a_i\right)^{2n-2}$$

is bounded above by a constant $\epsilon+1/(n-1)^{2n-2}$ when $a_i$ is sufficient large.

It is obvious that $$\frac{1}{(n-1)^{2n-2}}<\frac{2^n}{n^{2n-2}}$$ for every natural number $n>2$.

Hence we have complete the proof of step 1.

Step 2: The proof of step 1 implies that a sufficiently large number $M$ exists such that

$$\displaystyle \prod^n_{i=1} (1+a_i^2) /\left (\sum^n_{i=1} a_i\right)^{2n-2}<\frac{2^n}{n^{2n-2}}-\epsilon_1,\epsilon_1>0$$

outside the open box ${(a_1,\cdots,a_n),a_i<M,i=1,\cdots,n}$

Hence the solution to the constrainted optimization exists and lies inside the box.

We are trying to maximize $f (x_1,x_2, \ldots ,x_n)$ when $g (x_1,x_2, \ldots ,x_n)=0$

$$ f (x_1,x_2, \ldots ,x_n) = \log(1+x_1^2)+\log(1+x_2^2) \ldots +\log(1+x_n^2)-(2n-2)\log(x_1+x_2+\ldots +x_n)$$

$$g(x_1,x_2, \ldots ,x_n) = \log x_1+\log x_2 +\ldots +\log x_n $$

The existence of Lagrange multiplier suggest that $$\frac{\partial f}{\partial x_k}=\lambda\frac{\partial g}{\partial x_k}$$

i.e.,

$$\frac{2x_i}{1+x_i^2}-\frac{\lambda}{x_i}=\frac{2n-2}{\sum_{k=1}^n x_k},i=1,2,\cdots,n$$

If $x_i\neq x_j$ for some $i\neq j$,then

$$\lambda/2=\frac{(x_i x_j-1)x_i x_j}{(1+x_i^2)(1+x_j^2)}$$

Suppose that $x_i\neq x_j$ for $1\leq i<j\leq3$

Direct calculation shows that equations $$\lambda/2=\frac{(x_i x_j-1)x_i x_j}{(1+x_i^2)(1+x_j^2)},1\leq i<j\leq3$$ force some $x_i=x_j$ for $i\neq j$,a contradiction.

It is clear that $\{x_k\}$ has at most two elements.We have to solve the following equations

$$\frac{2x_i}{1+x_i^2}-\frac{\lambda}{x_i}=\frac{2n-2}{p x_1+q x_2},i=1,2,p+q=n,p>0,q>0,p,q\in\mathbb{Z},x_1\neq x_2$$

It suffices to solve $$\frac{2x_1^2}{1+x_1^2}-\frac{2x_2^2}{1+x_2^2}=\frac{(2n-2)(x_1-x_2)}{p x_1+q x_2}$$

which is equivalent to

$$(p+q-1)(1+x_1^2+x_2^2+x_1^2 x_2^2)=(p+q)x_1 x_2+px_1^2+qx_2^2$$

$LHS$ is larger than $2(p+q-1)x_1 x_2+px_1^2+qx_2^2$,a contradiction!

and the equations have only one solution,i.e.,$(1,1,\cdots,1)$.

Q.E.D.

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It is not quite clear what you mean by $a_i \to \infty$ in Step 1. Can the $a_i$ tend independently to $+\infty$ ? –  Ewan Delanoy Nov 17 '12 at 5:22
    
@EwanDelanoy:Yes,$a_i$ tends to infinity for a fixed index $i$. –  y zhao Nov 17 '12 at 6:34
    
Sorry, it's still not clear. Do you mean that only one $a_i$ moves while all the others stay fixed ? –  Ewan Delanoy Nov 18 '12 at 16:55
    
@EwanDelanoy:It suffices to see the fact that the product $A$ over $a_i>1$ and $\sum_{j=1}^n a_j$ tend to infinity when one $a_i$ tends to infinity,regardless of whether the others stay fixed or not. –  y zhao Nov 19 '12 at 4:12
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Let $l(a_1,\dots,a_n) = \prod\limits^n_{i=1} (1+a_i^2)$

Let $r(a_1,\dots,a_n) = \cfrac {2^n}{n^{2n-2}}\left (\sum\limits^n_{i=1} a_i\right)^{2n-2}$

$\cfrac{d^nl}{da_1\dots da_n}( a_1,\dots,a_n ) = 2 ^ n\prod\limits^n_{i=1} a_i$

$\cfrac{d^nr}{da_1\dots da_n}( a_1,\dots,a_n ) = \cfrac {2^n}{n^{2n-2}}\cfrac {(2n-2)!}{(n-2)!}(\sum\limits^n_{i=1} a_i)^{n-2}$

$(\sum\limits^n_{i=1} a_i)^{n} \ge n^n \prod\limits^n_{i=1} a_i$

$(\sum\limits^n_{i=1} a_i)^{n-2} \ge n^{n-2} (\prod\limits^n_{i=1} a_i)^{\frac{n-2}{n}}$

$\cfrac {2^n}{n^{2n-2}}\cfrac {(2n-2)!}{(n-2)!}(\sum\limits^n_{i=1} a_i)^{n-2} \ge \cfrac {2^n}{n^{2n-2}}\cfrac {(2n-2)!}{(n-2)!} n^{n-2} (\prod\limits^n_{i=1} a_i)^{\frac{n-2}{n}} = \cfrac {2^n}{n^{n}}\cfrac {(2n-2)!}{(n-2)!}(\prod\limits^n_{i=1} a_i)^{\frac{n-2}{n}}$

$1 \le \cfrac{1}{n^{n}}\cfrac {(2n-2)!}{(n-2)!}$

After this line, I assume $(\prod\limits^n_{i=1} a_i)^{\frac{n-2}{n}} = 1$ which I can't according to the comments.

$2^n \le \cfrac {2^n}{n^{n}}\cfrac {(2n-2)!}{(n-2)!} \le \cfrac {2^n}{n^{2n-2}}\cfrac {(2n-2)!}{(n-2)!}(\sum\limits^n_{i=1} a_i)^{n-2}$

$\cfrac{d^nl}{da_1\dots da_n}( a_1,\dots,a_n ) \le \cfrac{d^nr}{da_1\dots da_n}( a_1,\dots,a_n )$

$l( a_1,\dots,a_n ) \le r( a_1,\dots,a_n )$

$\prod\limits^n_{i=1} (1+a_i^2) \le \cfrac {2^n}{n^{2n-2}}\left (\sum\limits^n_{i=1} a_i\right)^{2n-2}$

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Derivatives of $f$ greater than that of $g$ do not imply $f$ is greater than $g$. If you want to integrate back to get the inequality, then you lose the condition $\prod a_i = 1$ when you deal with derivatives. (Which you used to derive an inequality about derivatives) –  Sanchez Nov 16 '12 at 20:01
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