Trig problem

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

i never tried to prove muirhead with AM-GM(had other jobs to do) but everything i could prove with muirhead i was also able to prove with AM-GM. let me think of it for a moment, maybe i can show you a proof of muirhead tomorrow(don't think i'll do that today).

Peter

**Posted:** Sun Jul 04, 2004 8:19 am

manlio

I am VERY VERY VERY interested in this equivalence between Muirhead and

AM-GM that I have never thought possible.

I have always considered AM-GM and Muirhead as two distinct statements so if

you can prove they are equivalent I will be astonished and very happy.

**Posted:** Sun Jul 04, 2004 8:35 am

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

don't hope too much, i may be absolutely wrong...i will try to proof while greece wins i hope Wink

(but it's not that bad if they don't...one student in my class must color his hair pink if portugal wins Mr. Green

)

Peter

**Posted:** Sun Jul 04, 2004 9:04 am

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

one can actually prove muirhead using AM-GM(or, to be precise the weighted version of it: if $\sum^n_{i=1}w_i=1$ , then $w_1a_1+...w_na_n\geq a_1^{w_1}\cdot...\cdot a_n^{w_n}$ : one can either prove it explicitly or use the normal AM-GM to prove it for rational $w_i$ 's, and then extend to irrational by continuity).
so, let's proof muirhead by induction on the number n of variables.
n=2: nearly trivial(let the exponents be $x_1,x_2$ , $y_1,y_2$ where $x_1\geq y_1\geq y_2\geq x_2$ and $x_1+x_2=y_1+y_2$ ). then there exist $w_1,w_2$ such that $0\leq w_1,w_2\leq 1$ , $w_1+w_2=1$ and $y_1=w_1x_1+w_2x_2$ , thus we also have $y_2=w_2x_1+w_1x_2$ . let a,b be arbitrary positive reals. then $\displaystyle a^{x_1}b^{x_2}+a^{x_2}b^{x_1}=(w_1a^{x_1}b^{x_2}+w_2a^{x_2}b^{x_1})+(w_2a^{x_1}b^{x_2}+w_1a^{x_2}b^{x_1})\geq a...$ .
(n-1)->n: let $x_1,...,x_n$ and $y_1,...y_n$ be the exponents with the muirhead conditions. since $x_1\geq y_1$ and $x_n\leq y_n$ , there is some k such that $x_k\geq y_k$ , $x_{k+1}\leq y_{k+1}$ . then by the case n=2, we may decrease $x_k$ and increase $x_{k+1}$ such that their sum is invariant until either $x_k=y_k$ or $x_{k+1}=y_{k+1}$ . it is clear that the muirhead conditions are still satisfied. so suppose two of the exponents are equal, $x_k=y_k$ . now consider those terms of the symmetric sum where the k-th exponent has the same variable. then using muirhead for the other n-1 variables(the other n-1 exponents satisfy the muirhead conditions as well), we get the desired inequality.

it is clear that muirhead implies AM-GM.

(btw: two true statements are always equivalent Mr. Green

)

Peter

**Posted:** Sun Jul 04, 2004 11:03 am

darij grinberg

Peter, I think your induction step needs a few modifications. For instance, when you say that "we may decrease $x_k$ and increase $x_{k+1}$ such that their sum is invariant until either $x_k=y_k$ or $x_{k+1}=y_{k+1}$ . it is clear that
the muirhead conditions are still satisfied.", I think you make an additional assumption, namely that for every $i \leq k$ , we have $x_i \geq y_i$ . Hence, it is not enough to say "there is some k such that $x_k\geq y_k$ , $x_{k+1}\leq y_{k+1}$ ", but you also must point out that you take the first such k.

Then I have some little notes on your last argument, when you use the n-1 assumption. I'll post all this later.

Meanwhile, I just wanted to note that you don't need any AM-GM for n = 2. In fact, from the conditions $x_1 \geq y_1 \geq y_2 \geq x_2$ and $x_1 + x_2 = y_1 + y_2$ , it follows that $y_1-x_2\geq 0$ and $y_2-x_2\geq 0$ , so that $b \geq a$ entails $b^{y_{1}-x_{2}}-a^{y_{1}-x_{2}} \geq 0$ and $b^{y_{2}-x_{2}}-a^{y_{2}-x_{2}} \geq 0$ , while $b \leq a$ yields $b^{y_{1}-x_{2}}-a^{y_{1}-x_{2}} \leq 0$ and $b^{y_{2}-x_{2}}-a^{y_{2}-x_{2}} \leq 0$ . In both cases, we have $\left( b^{y_{1}-x_{2}}-a^{y_{1}-x_{2}}\right) \left( b^{y_{2}-x_{2}}-a^{y_{2}-x_{2}}\right) \geq 0$ . Now,

$0\leq a^{x_{2}}b^{x_{2}}\left( b^{y_{1}-x_{2}}-a^{y_{1}-x_{2}}\right) \left( b^{y_{2}-x_{2}}-a^{y_{2}-x_{2}}\right)$
$=a^{x_{2}}b^{x_{2}}\left( b^{y_{1}+y_{2}-2x_{2}}-b^{y_{1}-x_{2}}a^{y_{2}-x_{2}}-a^{y_{1}-x_{2}}b^{y_{2}-x_{2}}+a^{y_{1}+y_{2}...$
$=a^{x_{2}}b^{x_{2}}\left( b^{x_{1}-x_{2}}-b^{y_{1}-x_{2}}a^{y_{2}-x_{2}}-a^{y_{1}-x_{2}}b^{y_{2}-x_{2}}+a^{x_{1}-x_{2}}\right...$
$=a^{x_{2}}b^{x_{1}}-a^{y_{2}}b^{y_{1}}-a^{y_{1}}b^{y_{2}}+a^{x_{1}}b^{x_{2}}$ .

Darij

**Posted:** Mon Jul 05, 2004 4:21 am

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Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

[quote="darij grinberg"] I think you make an additional assumption, namely that for every $i \leq k$ , we have $x_i \geq y_i$ . [/qoute]

i don't think so. actually, since their sum is invariant, we don't have problems for larger smaller indices where only their sum influences the inequalities of muirhead and for larger indices, these terms don't play any role in the muirhead conditions. but maybe my mind is a bit blocked.
to your proof of muirhead for n=2: that is just the proof for AM-GM for n=2. muirhead is in my opinion that easy to proof that one can't say that it is "equivalent" to some other inequality.

what are your comments to using the inequality for n-1 variables?

Peter

**Posted:** Mon Jul 05, 2004 5:20 am

manlio

Dear Peter , I have studied deeply your proof but it is not clear when you say:

....then by the case n=2 we may decrease x_k and increase x_k+1 such that their sum is invariant until either x_k=y_k or x_k+1=y_k+1. It is clear that the muirhead conditions are still satisfied....

can you please detail better this fundamental step as I want to understand entirely your proof which seems very interesting.

Thank you very much.

**Posted:** Mon Jul 05, 2004 8:22 am

darij grinberg

**Posted:** Mon Jul 05, 2004 9:58 am

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Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

i do not need that assertion, darij. actually, i wrote down my muirhead conditions another way(it is equivalent): $x_n\leq y_n$ , $x_{n-1}+x_n\leq y_{n-1}+y_n$ , ..., $x_1+...+x_n=y_1+...+y_n$ where $x_1\geq x_2\geq ... \geq x_n$ and $y_1\geq y_2\geq ... \geq y_n$ . now for indices $i\leq k$ these conditions are sure still satisfied since the sum is invariant. for $i>k+1$ it is also clear since we change nothing. so look at $x_{k+1}+...+x_n\leq y_{k+1}+...+y_n$ . we have $x_{k+2}+...x_n\leq y_{k+2}+...y_n$ and $x_{k+1}\leq y_{k+1}$ (after the increasion of $x_{k+1}$ as well, since we stop when we have inequality), so that's clear. it is clear as well that $x_1\geq x_2\geq ... \geq x_n$ is still satisfied since $x_k\geq y_k\geq y_{k+1}\geq x_{k+1}$ . so i think there are no problems. but i could be wrong Sad

Peter

**Posted:** Mon Jul 05, 2004 10:20 am

Peter Scholze · Yang-Mills Theory Offline Joined: 03 Jun 2004 Posts: 674

for manlio: i may decrease $x_k$ and increase $x_{k+1}$ such that their sum is invariant because looking at the symmetric sum where the other exponents have the same variables and the k-th and k+1-th exponent change their variables, we may use the muirhead for n=2. that gives that these terms are greater than when decreasing $x_k$ and increasing $x_{k+1}$ , so it suffices to show the inequality for the case when $x_k=y_k$ or $x_{k+1}=y_{k+1}$ , where we can use muirhead for n-1 variables.
example(i think you understand what i mean, but maybe it helps to understand the idea): we want to prove $\sum a^3b^3\geq \sum a^2b^2c^2$ . now, after muirhead for n=2, we have $b^3+c^3\geq b^2c+c^2b$ , but the left side is $a^3(b^3+c^3)+b^3(a^3+c^3)+c^3(a^3+b^3)\geq a^3b^2c+a^3bc^2+b^3a^2c+b^3ac^2+c^3a^2b+c^3ab^2=\sum a^3b^2c$ . but after muirhead for n-1=2 variables, we have $a^3c+ac^3\geq a^2c^2+a^2c^2$ , which gives the desired inequality.
basically, we are just moving the weights down the variables to get $\{x_k\}$ to be the same as $\{y_k\}$ where moving the weights is justified with AM-GM.

Peter

**Posted:** Mon Jul 05, 2004 10:29 am

darij grinberg

[Edit: I have slightly simplified an easy step now.]

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galois · Riemann Hypothesis Offline Joined: 11 Jul 2003 Posts: 394

hey i think there is aproof of muirhead's theorem in kedlaya's packet on inequalities and it seems to be shorter than dg's monstrous version Mr. Green

(no offence intended mate)

**Posted:** Mon Jul 05, 2004 10:55 am

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darij grinberg

**Posted:** Mon Jul 05, 2004 10:59 am

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fuzzylogic · Yang-Mills Theory Offline Joined: 24 Mar 2004 Posts: 719

The proof of Muirhead in Kiran Kedlaya's Inequality notes is very short, but it relies on a fact which he didn't include a proof.

If $s$ majorizes $t$ , then there exist nonnegative constants $k_{\sigma}$ , as $\sigma$ ranges over the permutations of $\{1, \dots, n\}$ , whose sum is 1 and such that

$\sum_\sigma k_{\sigma} (s_{\sigma(1)}, \dots, s_{\sigma(n)}) = (t_1, \dots, t_n)$

Armed with this fact, the proof of Muirhead is almost trivial by weighted AM-GM.

This fact can also be used to prove the Majorization Inequality (aka Karamata Inequality).

**Posted:** Mon Aug 02, 2004 2:34 pm

perfect_radio

Does anyone know a proof for the lemma Kedlaya used in proving Muirhead?

(i recently downloaded his ineq notes and i was quite interested in the proof of muirhead; the one given here seems pretty simple, but i want to know more about this other approach)

**Posted:** Thu Aug 04, 2005 6:02 pm

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