easy inequality

mestav · P versus NP Offline Joined: 14 Dec 2008 Posts: 22

Let $a,b,c,d$ are real numbers

$(4,0,0,0) + 3(2,2,0,0) \geq 4(3,1,0,0)$ or

$3(a^{4} + b^{4} + c^{4} + d^{4}) + 6(a^{2}b^{2} + a^{2}c^{2} + a^{2}d^{2} + b^{2}c^{2} + b^{2}d^{2} + c^{2}d^{2}) \geq 4(a^{3...$

aadil

what do you mean by the first inequality?

**Posted:** Sat Nov 07, 2009 3:54 am

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**Posted:** Sat Nov 07, 2009 4:39 am

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mestav · P versus NP Offline Joined: 14 Dec 2008 Posts: 22

yes but not cyclic it's symmetric

**Posted:** Sat Nov 07, 2009 4:58 am

Guest · Hodge Conjecture Offline Joined: 18 May 2005 Posts: 79

**Posted:** Sat Nov 07, 2009 4:58 am

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earldbest

**Posted:** Sat Nov 07, 2009 4:59 am

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mestav · P versus NP Offline Joined: 14 Dec 2008 Posts: 22

original inequality is $a^{3}(b + c + d) + b^{3}(a + c + d) + c^3(a + b + d) + d^3(a + b + c)\le\frac {3}{4}(a^{2} + b^{2} + c^{2} + d^{2})^2$

I multiplied and I have this inequality

is original inequality false? or my operations false?

**Posted:** Sat Nov 07, 2009 5:18 am

dgreenb801

It's equivalent to $(a - b)^4 + (a - c)^4 + (a - d)^4 + (b - c)^4 + (b - d)^4 + (c - d)^4 \ge 0$ , but I think the right side of the inequality should be $4(a^3b + b^3a + a^3c + c^3a + a^3d + d^3a + b^3c + c^3b + b^3d + d^3b + c^3d + d^3c)$ instead of what you have.

**Posted:** Sat Nov 07, 2009 3:30 pm

mestav · P versus NP Offline Joined: 14 Dec 2008 Posts: 22

I am sorry I fixed it and thank you dgreenb801

**Posted:** Sun Nov 08, 2009 2:38 am

spanferkel · Yang-Mills Theory Offline Joined: 15 Oct 2005 Posts: 638

The convention in this forum is to use square brackets for symmetrical sums, like $[3,1,0,0]$ . Note that this sum has always $n!$ terms, e.g. $[4,0,0,0] = 6\sum a^4$ , as the symmetrical sum counts all the permutations.

If one writes $\sum_{sym}$ , it is understood to count each different term only once, e.g. if we deal with 4 variables, $\sum_{sym}a^3b$ has 12 terms, not 24. And $[3,1,0,0]=2\sum_{sym}a^3b$ .

This sometimes leads to confusion, let's just be coherent in each line. Smile

**Posted:** Sun Nov 08, 2009 8:11 am

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