Very Hard and very Beautiful Harazi Inequality

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Very Hard and very Beautiful Harazi Inequality

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Source: Harazi

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manlio

#1 h Dec 22, 2004, 7:26 PM • 2 Y

Y by Adventure10, Mango247

For $x_i$ , $i=1,...,n$ , positive real numbers such that

$\frac{1}{x_1}+...+\frac {1}{x_n}=n$ prove the inequality

$\displaystyle x_1\cdot...\cdot x_n-1\geq (\frac{n-1}{n})^{n-1}\cdot (x_1+x_2+...x_n-n)$

Really beautiful

Like source it was better to write: "Harazi genial mind" but I didn't know if Harazi was agree with me, so I didn't write it.

This post has been edited 1 time. Last edited by manlio, Dec 23, 2004, 10:45 AM

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#2 h Dec 22, 2004, 7:30 PM • 2 Y

Y by Adventure10, Mango247

Do you mean to have negative terms on both sides?
Or should the reciprocals perhaps sum to $n$ ?

Edit: He got there first!

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Vasc

#3 h Dec 23, 2004, 9:24 AM • 2 Y

Y by Adventure10, Mango247

Excellent inequality!
From here, the following statement follows:
If $x_1,x_2,...,x_n$ are positive numbers such that $\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}=n$ then
$\displaystyle \frac{x_1+x_2+...+x_n-n}{x_1x_2...x_n-1}<e$

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harazi

#4 h Dec 23, 2004, 10:33 AM • 2 Y

Y by Adventure10, Mango247

Manlio, I thought you were among those who said that we must have smart topic titles. The title of this topic does not satisfy this criteria! By the way, Vasc's inequality will not remain unsolved for months if someone proves this inequality!

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manlio

#5 h Dec 23, 2004, 10:54 AM • 2 Y

Y by Adventure10, Mango247

You are right Harazi, but for this inequality I made an exception as it is too much nice, fantastic and superlative

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#6 h Dec 23, 2004, 11:27 AM • 2 Y

Y by Adventure10, Mango247

is it possible that $(\frac{n-1}{n})^{n-1}$ is not best possible? at least, for $n=3$ , we might as well use $\frac 76$ ...
i'm asking just because i don't see how a proof could use the constant $(\frac{n-1}{n})^{n-1}$ ...but i didn't solve a hard inequality for times...

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Vasc

#7 h Dec 23, 2004, 11:45 AM • 2 Y

Y by Adventure10, Mango247

I found that $\displaystyle (\frac{n-1}{n})^{n-1}$ is the best constant.

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harazi

#8 h Dec 23, 2004, 2:41 PM • 2 Y

Y by Adventure10, Mango247

Me too!

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#9 h Dec 23, 2004, 3:26 PM • 2 Y

Y by Adventure10, Mango247

sorry for doing nonsense...of course it's best possible. indeed, i only saw that $\frac 7{18}$ works which is worse than $\frac 49$ .

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#10 h Dec 23, 2004, 5:13 PM • 2 Y

Y by Adventure10, Mango247

now i found a proof

however, it's too ugly that i don't want to disturb you with it

its main idea is too take the minimum on some bounded region and then obtain some properties of it by replacing $x_1,x_2$ , $x_1>x_2$ with $\frac{x_1}{1+\epsilon x_1},\frac{x_2}{1-\epsilon x_2}$ , respectively. by some ugly arguments, we then see that there are two groups $x_1=...=x_k$ and $x_{k+1}=...=x_n$ . it is then easy to see that we may assume $k=1$ and from that point on, it's easy.

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manlio

#11 h Dec 29, 2004, 11:33 AM • 2 Y

Y by Adventure10, Mango247

Harazi said kindly to me that the solution is very very hard and very very long and it uses induction + mixing variables technique.

Unfortunately Harazi has no time to post his solution because he is very busy with his studies, so I would like to know if anyone else is very interested in it and could help me to find a solution.

Unfortunately I am not a good solver of inequalities but I have a lot of time to spend on it.

So if you are interested in solving it and to collaborate with me, please contact me.

Thank you very much and Happy New Year.

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#12 h Dec 29, 2004, 2:49 PM • 2 Y

Y by Adventure10, Mango247

so, here's my proof:

$n=2$ is trivially gives equality, so let's assume $n>2$ .

let's consider first $x_i\in [a;b]$ for some interval $[a;b]$ . then the $x_i$ are in a closed and bounded domain, therefore the minimum of $f(x_1,...,x_n)=x_1\cdots x_n - 1 - (\frac{n-1}{n})^{n-1}(x_1+...+x_n-n)$ is assumed for some $x_1,...,x_n$ . we will now look at what happens if we slightly change the variables.

assume that $x_i>x_j$ . set $y_k=x_k$ for $k\neq i,j$ , $y_i=\frac{x_i}{1+\epsilon x_i}$ , $y_j=\frac{x_j}{1-\epsilon x_j}$ . one easily checks that the condition $\sum \frac{1}{y_i}=n$ is still satisfied.

we do now calculate the derivative of $g(\epsilon)=f(y_1,...,y_n)$ at $\epsilon=0$ . note that $y_i$ is about $x_i-\epsilon x_i^2$ and $y_j$ is about $x_j+\epsilon x_j^2$ for $\epsilon\to 0$ ; therefore we may use these estimates when calculating $g'(0)$ . the result is after a simple calculation
$g'(0)=(x_j-x_i)(x_1\cdots x_n-(\frac{n-1}{n})^{n-1}(x_j+x_i))$

for small $\epsilon>0$ , $y_1,...,y_n$ stays in the bounded region; therefore $f(y_1,...,y_n)\geq f(x_1,...,x_n)$ ; this implies $g'(0)\geq 0$ which by the above gives
$x_1\cdots x_n\leq (\frac{n-1}{n})^{n-1}(x_j+x_i)$ (1)
furthermore, if $a<x_2<x_1<b$ , then we have equality since $y_1,...,y_n$ stay in the bounded region for small $\epsilon<0$ as well.

now, we see that if we have equality in (1), then we may replace $x_i,x_j$ by arbitrary $y_i,y_j$ with equal harmonic mean after a short calculation. therefore we may assume that all terms which lie between $a$ and $b$ are equal.

now, the above holds for all intervals $[a; b]$ . we will take intervals $[\frac{1}{b}; b]$ and study what happens as $b\to\infty$ . our goal is to show that we only have to show the inequality for $n-1$ out of the $n$ variables equal.

let $(x_1)_b\leq ...\leq (x_n)_b$ be minimal for the interval $[\frac{1}{b}; b]$ . we only consider $b>n$ ; in that case, we may not have $(x_1)_b=a=\frac{1}{b}$ since we have $\sum (x_i)_b=n$ . therefore we immediately see that $\frac{1}{b}<(x_1)_b=...=(x_{l_b-1})_b<(x_{l_b})_b=...=(x_n)_b=b$ for some $l_b$ by the above.

assume that $(x_{n-1})_b=(x_n)_b=b$ for some large $b$ . for $b>1$ , not all variables may be equal, therefore $(x_1)_b<(x_n)_b$ . but we have $(x_1)_b \cdots (x_{n-2})_b\geq (\frac{n-2}{\frac 1{(x_1)_b}+...+\frac 1{(x_{n-2})_b}})^{n-2}=(\frac{n-2}{n-\frac 2b})^{n-2}\geq (\frac{n-2}{n})^{n-2}>e^{-2}$ for $n>2$ . the LHS of (1) is thus at least $e^{-2}b^2$ , the RHS is at most $n(\frac{n-1}{n})^{n-1}(b-1)$ . for large $b$ , inequality (1) therefore fails, and we get that for all large $b$ , $(x_{n-1})_b<(x_{n})_b$ if $(x_n)_b=b$ .

if $(x_n)_b<b$ , then all variables would be equal; but that case is trivial. thus we only have to consider that case that $l_b=n$ :
$\frac{1}{b}<(x_1)_b=...=(x_{n-1})_b<(x_n)_b=b$ ; thus $n-1$ out of the $n$ variables are equal. furthermore, we may assume that one variable is large as we need it; in our argumentation here we may take $b$ as large as we want and always get the optimal solution if not the solution with $x_1=...=x_n$ where the inequality trivially holds to be the one with $(x_n)_b=b$ .

now, the rest are simple explicit calculations. we have $x_n=b$ , but $n=\frac{n-1}{x_1}+\frac{1}{x_n}=\frac{n-1}{x_1}+\frac{1}{b}$ , so we get $x_1=...=x_{n-1}=\frac{n-1}{n-\frac 1b}$ . then we get

$L=x_1\cdots x_n-1=b(\frac{n-1}{n-\frac 1b})^{n-1}-1$ ,
$R=(\frac{n-1}{n})^{n-1}(x_1+...+x_n-n)=(\frac{n-1}{n})^{n-1} \frac{-2+b+\frac 1b}{1-\frac 1{nb}}$ .

thus $L\geq R$ iff
$b(\frac{n}{n-\frac 1b})^{n-1}-(\frac{n}{n-1})^{n-1}\geq \frac{nb^2-2nb+n}{nb-1}$ .

but we know that $(1+\frac{\frac 1b}{n-\frac 1b})^{n-1}\geq 1+\frac{n-1}{nb-1}$ by bernoulli and let $(\frac{n}{n-1})^{n-1}=:p$ . we get that $L\geq b+\frac{nb-b}{nb-1}-p=\frac{nb^2-2nb+((3-p)nb-2b+p)}{nb-1}$ . we only need to have $((3-p)n-2)b+p\geq n$ . since we may take $b$ as large as we want, it suffices that $(3-p)n>2$ . now, it's a simple check that this indeed holds for all $n$ (since $p<e$ for all $n$ , we immediately get the result for $n>7$ and the rest can be done by hand).

i hope there's no mistake in here

Peter

This post has been edited 11 times. Last edited by Peter Scholze, Feb 27, 2005, 1:42 PM

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#13 h Dec 29, 2004, 4:18 PM • 2 Y

Y by Adventure10, Mango247

hmm there's a flaw in the argument that $l_b=n$ ...wait a moment...

Edit: i hope it's correct now

This post has been edited 1 time. Last edited by Peter Scholze, Dec 29, 2004, 5:02 PM

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manlio

#14 h Dec 29, 2004, 4:23 PM • 2 Y

Y by Adventure10, Mango247

Thank you very much, Peter.

I will study deeply it.

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manlio

#15 h Feb 10, 2005, 9:05 PM • 2 Y

Y by Adventure10, Mango247

Harazi hint : prove this one by mixing variables + induction.

Two questions for you Harazi.

Do you use induction method and then mixing variables to prove induction step or do you use mixing variables method and then induction to prove it?

Do you use a particular substitution for x_i?

Thank you very much.

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#16 h Feb 27, 2005, 1:50 PM • 2 Y

Y by Adventure10, Mango247

i would be glad if anyone could check whether my proof is correct... i asked manlio and darij to check it but darij never got through the proof and manlio never answered

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manlio

#17 h Feb 27, 2005, 2:43 PM • 2 Y

Y by Adventure10, Mango247

I really read your proof many times, your reasoning is very nice and very original, but very hard to follow.

At first sight, in my opinion, it seems good but I would like to check it again.

Harazi solved it in a more simple way using mixing variables theory, but he never posted his solution and ,until now, nobody can reconstruct Harazi reasoning.

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harazi

#18 h Feb 27, 2005, 2:45 PM • 2 Y

Y by Adventure10, Mango247

Not even me.

I simply forgot my reasoning which I had when creating it. I will try to solve it now.

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manlio

#19 h Feb 27, 2005, 2:50 PM • 2 Y

Y by Adventure10, Mango247

Thank you very much, Harazi.

It is really a super-beautiful inequality (in my opinion).

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5526 posts

harazi

#20 h Feb 27, 2005, 2:53 PM • 2 Y

Y by Adventure10, Mango247

Manlio, please, I became alergic to words like super-beautiful. It is just a nice inequality that I cannot prove for momment. That's all.

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Soarer

#21 h Aug 13, 2005, 9:50 PM • 2 Y

Y by Adventure10, Mango247

Very strange thing I have found. Where's the flaw?
To prove:
If $\sum \frac{1}{x_1} = n$ , then $x_1...x_n - 1 \ge (\frac{n-1}{n})^{n-1} (x_1+...+x_n-n)$

First, we can replace $x_1$ by $\frac{1}{x_1}$ , then we have the equivalent inequality,
$\sum x_1 = n$ implies $\frac{1}{x_1...x_n} - 1 \ge (\frac{n-1}{n})^{n-1}(\frac{1}{x_1}+...+\frac{1}{x_n}-n)$
Now homogenize it we have
$\frac{(\frac{x_1+...+x_n}{n})^n}{x_1...x_n} - 1 \ge (\frac{n-1}{n})^{n-1}((\frac{1}{x_1}+...+\frac{1}{x_n})(\frac{x_1+...+x_n}{n})-n)$
Dehomogenize by assuming $\sum \frac{1}{x_1} = n$ , we have
$\frac{(\frac{x_1+...+x_n}{n})^n}{x_1...x_n} - 1 \ge (\frac{n-1}{n})^{n-1}(x_1+...+x_n-n)$

That means when $\sum \frac{1}{x_1} = n$ , $A:x_1...x_n - 1 \ge (\frac{n-1}{n})^{n-1} (x_1+...+x_n-n)$ if and only if $B:\frac{(\frac{x_1+...+x_n}{n})^n}{x_1...x_n} - 1 \ge (\frac{n-1}{n})^{n-1}(x_1+...+x_n-n)$ .
Thus we only need to prove, $A+B:x_1...x_n - 1 + \frac{(\frac{x_1+...+x_n}{n})^n}{x_1...x_n} - 1 \ge (\frac{n-1}{n})^{n-1} (x_1+...+x_n-n) + (\frac{n-1}{n})^{n-1}(x_1+...+x_n-n)$
because if there exists some n-tuples such that A is true, then B is also true and vice versa. If there exists some n-tuples such that A is reversed, then B is also reversed and vice versa. Thus we have to prove $A+B$ is true under the condition $\sum \frac{1}{x_1} = n$

But this is easy because $(\frac{n-1}{n})^{n-1} \le \frac{1}{2}$ so R.H.S. $\le (x_1+...+x_n-n)$ while L.H.S. $= \frac{(\frac{x_1+...+x_n}{n})^n}{x_1...x_n} + x_1...x_n + n - 2 - n \ge x_1+...+x_n - n$ .

I think this is completely fallacious, but I simply can't figure out where I am wrong..

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harazi

#22 h Aug 14, 2005, 6:21 AM • 2 Y

Y by Adventure10, Mango247

I don't understand your argument with $A,B$ and $A+B$ and my logic tells me it's wrong.

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Soarer

#23 h Aug 14, 2005, 7:21 AM • 1 Y

Y by Adventure10

harazi wrote:

I don't understand your argument with $A,B$ and $A+B$ and my logic tells me it's wrong.

I'm not sure about that part either.
but is it $A \ge 0$ and $B \ge 0$ occurs at hte same time?

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#24 h Aug 14, 2005, 9:43 AM • 2 Y

Y by Adventure10, Mango247

you only proved that ( $A$ is true for all $x_1,...,x_n$ ) <=> ( $B$ is true for all $x_1,...,x_n$ ). but now, if $x_1,...,x_n$ is a counterexample to $A$ it doesn't mean it is one for $B$ - it could even be true that $B$ is very unsharp for $x_1,...,x_n$ , so that in the sum, $A+B$ is still true.

if you proved that, given $x_1,...,x_n$ , ( $A$ is true for $x_1,...,x_n$ ) <=> ( $B$ is true for $x_1,...,x_n$ ), then your argument worked.

btw, your argument worked as well for $\left(\frac{n-1}{n}\right)^{n-1}$ replaced by $\frac 12$ - but, as already mentioned, $\left(\frac{n-1}{n}\right)^{n-1}$ is best possible.

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Soarer

#25 h Aug 14, 2005, 10:20 AM • 2 Y

Y by Adventure10, Mango247

i see. Thanks!

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#26 h Feb 8, 2006, 1:56 PM • 2 Y

Y by Adventure10, Mango247

Peter Scholze wrote:

now, the above holds for all intervals $[a; b]$ . we will take intervals $[\frac{1}{b}; b]$ and study what happens as $b\to\infty$ . our goal is to show that we only have to show the inequality for $n-1$ out of the $n$ variables equal.

Peter

Sorry,it was long ago.
Hi,Peter Scholze,I can't follow you solution,this is the first place that I can't understand,can you explain more,I am sure that your idea must be nice.

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#27 h Feb 8, 2006, 2:11 PM • 2 Y

Y by Adventure10, Mango247

Anyway,here is my solution:
Let $y_i=\frac 1{x_i},i=1,2,\cdots,n$
We have $\sum_{i=1}^ny_i=n$
we should to prove:
$\frac{1}{y_1\cdots y_n}-1 \ge (\frac{n-1}{n})^{n-1}(\sum_{i=1}^n\frac{1}{y_i}-n)$
Let $f(y_1,y_2,\cdots,y_n)=\frac{1}{y_1\cdots y_n}- (\frac{n-1}{n})^{n-1} \sum_{i=1}^n\frac{1}{y_i}$
We have: $f(y_1,y_2,\cdots,y_n) \ge f(\frac{y_1+y_2}2,\frac{y_1+y_2}2,\cdots,y_n) \iff (y_1+y_2)y_3\cdots y_n \le (\frac n{n-1})^{n-1}$
clearly true.
Then use mixing varialbes.
I hope my solution is correct and nice

Anyway,Peter Scholze can you explain your idea,because I want to know your nice solution,too.Thanks

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