AM-GM for 3 variables

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Eugenis

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Eugenis

#1 h Aug 18, 2015, 1:04 PM • 2 Y

Y by Adventure10, Mango247

Prove $\dfrac{a^3+b^3+c^3}{3}\geq abc$ without using AM-GM for $a, b, c \ge 0.$

This post has been edited 3 times. Last edited by Eugenis, Aug 18, 2015, 1:09 PM

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jlammy

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#2 h Aug 18, 2015, 1:06 PM • 2 Y

Y by Adventure10, Mango247

Trivial

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Reason: added hide tags

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rkm0959

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rkm0959

#3 h Aug 18, 2015, 1:08 PM • 2 Y

Y by Adventure10, Mango247

Note that $\frac{a^3+b^3+c^3}{3} \ge abc \iff a^3+b^3+c^3-3abc \ge 0 \iff (a+b+c)((a-b)^2+(b-c)^2+(c-a )^2) \ge 0$
From $a+b+c > 0$ and the trivial inequality, we are done.

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PiOfLife314

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PiOfLife314

#4 h Aug 18, 2015, 1:08 PM • 3 Y

Y by Porifera, Adventure10, Mango247

jlammy wrote:

By AM-GM

Eugenis wrote:

without AM-GM.

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jlammy

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jlammy

#5 h Aug 18, 2015, 1:08 PM • 2 Y

Y by Adventure10, Mango247

when I posted, there was the condition that $a,b,c$ positive.

screw you Eugenis!

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Eugenis

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Eugenis

#6 h Aug 18, 2015, 1:09 PM • 2 Y

Y by Adventure10, Mango247

Yeah I changed the problem statement after jlammy posted; sorry!

Also, assume $a, b, c \ge 0.$

@above Darn I'm screwed!

This post has been edited 1 time. Last edited by Eugenis, Aug 18, 2015, 1:10 PM

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PiOfLife314

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PiOfLife314

#7 h Aug 18, 2015, 1:09 PM • 2 Y

Y by Adventure10, Mango247

jlammy wrote:

when I posted, there was the condition that $a,b,c$ positive.

Oh, okay. Sorry!

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Eugenis

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Eugenis

#8 h Aug 18, 2015, 1:11 PM • 2 Y

Y by Adventure10, Mango247

Just a question: AM-GM for 3 variables can be proven by cubing both sides and subtracting the RHS right?

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ThE-dArK-lOrD

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ThE-dArK-lOrD

#9 h Aug 18, 2015, 1:11 PM • 2 Y

Y by Adventure10, Mango247

For $a,b,c \in \mathbb{R}^{+}$
$a^3+b^3+c^3-3abc=(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)\frac{1}{2} \geq 0$
So $\frac{a^3+b^3+c^3}{3} \geq abc$

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Eugenis

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Eugenis

#10 h Aug 18, 2015, 1:12 PM • 2 Y

Y by Adventure10, Mango247

Dark-Lord, how did you get the factorization? Can you explain that?

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rkm0959

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rkm0959

#11 h Aug 18, 2015, 1:13 PM • 2 Y

Y by Adventure10, Mango247

$a^3+b^3+c^3-3abc = a^3+b^3+3ab(a+b)+c^3-3ab(a+b+c) =$
$(a+b)^3+c^3-3ab(a+b+c)=(a+b+c)((a+b)^2-(a+b)c+c^2)-3ab(a+b+c)=$
$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

This post has been edited 2 times. Last edited by rkm0959, Aug 18, 2015, 1:15 PM

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Eugenis

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Eugenis

#12 h Aug 18, 2015, 1:16 PM • 2 Y

Y by Adventure10, Mango247

I don't see the motivation for factoring like that...

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jlammy

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jlammy

#13 h Aug 18, 2015, 1:18 PM • 2 Y

Y by Adventure10, Mango247

$\begin{vmatrix} a&b&c \\ c&a&b \\ b&c&a \end{vmatrix}=a^3+b^3+c^3-3abc$ . This can help us factorise.

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ThE-dArK-lOrD

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ThE-dArK-lOrD

#14 h Aug 18, 2015, 1:19 PM • 2 Y

Y by Adventure10, Mango247

Eugenis wrote:

I don't see the motivation for factoring like that...

It's the well-known factorization.

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Not_a_Username

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Not_a_Username

#15 h Aug 18, 2015, 1:21 PM • 2 Y

Y by Adventure10, Mango247

Usually, when factoring multivariable polynomials, we usually set $a=\pm b, a=\pm c, a=\pm (b+c)$ , and $a=b-c$ to try to factor the expression. In this case, $a=-b-c$ works out.

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Eugenis

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Eugenis

#16 h Aug 18, 2015, 1:23 PM • 2 Y

Y by Adventure10, Mango247

Wait but then how do you factor a 2 degree polynomial then?

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Not_a_Username

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Not_a_Username

#17 h Aug 18, 2015, 1:30 PM • 2 Y

Y by Adventure10, Mango247

I was talking about factoring $a^3+b^3+c^3-3abc$ . To factor $a^2+b^2+c^2-ab-bc-ca$ , you complete the square.

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Eugenis

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Eugenis

#18 h Aug 18, 2015, 1:31 PM • 2 Y

Y by Adventure10, Mango247

Wait I'm still a bit unclear how to factors $a^3+b^3+c^3-3abc$ .

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champion999

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champion999

#19 h Aug 18, 2015, 1:50 PM • 3 Y

Y by Cube1, Adventure10, Mango247

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$

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Cube1

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Cube1

#20 h Aug 18, 2015, 2:10 PM • 1 Y

Y by Adventure10

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)=(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$

So Eugenis, you are given that $a, b, c \ge 0$ so adding nonnegative terms will still make it nonnegative
And the square of any term is always nonnegative, so when we add nonnegative terms, we will again still make it nonnegative

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jlammy

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jlammy

#21 h Aug 18, 2015, 2:18 PM • 2 Y

Y by champion999, Adventure10

Let's overkill this:

Theorem (Dedekind). Let ${Xg}$ be a set of formal variables indexed by elements of a finite abelian group $G$ , and define $\phi(G):=\det \left (X_{gh^{-1}} \right)$ to be the group determinant. Then we have the factorization $\phi(G)=\prod_{\chi \in \hat{G}} \left ( \sum_{g\in G}\chi (g)X_g \right ),$ where $\hat{G}$ is the character group of $G$ , i.e. the group of homomorphisms from $G \to \mathbb{C}^{\times}$ .

For a proof, see this article. Taking $G=\mathbb{Z}/3\mathbb{Z}$ , $\phi(G)$ is equivalent to the circulant $\begin{vmatrix} a&b&c \\ c&a&b \\ b&c&a \end{vmatrix}=(a+b+c)(a+\omega b + \omega^2 c)(a+\omega^2 b + \omega c),$ where $\omega = \mathrm{cis} \left (\tfrac{2\pi}{3} \right )$ , the third root of unity.

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rkm0959

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rkm0959

#22 h Aug 18, 2015, 2:23 PM • 3 Y

Y by Eugenis, Adventure10, Mango247

Theorem 3 in http://www.artofproblemsolving.com/community/c3h532566p3054176 helps as well.
Let's factorize $a^3+b^3+c^3-3abc$ using Theorem 3.
From Theorem 3, we realize that $a^3+b^3+c^3-3abc = m(a+b+c)^3+n(a+b+c)(ab+bc+ca)+kabc$
Now setting $(a,b,c)=(2,-1,-1),(1,0,0),(1,1,1)$ gives that $m=1, n=-3$ .
$a^3+b^3+c^3-3abc=(a+b+c)^3-3(a+b+c)(ab+bc+ca) = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ .
Another trick for 3-var factorization is that:
If $f(x,y,z)=f(y,z,x)=f(z,x,y)=-f(x,z,y)$ for all $x,y,z$ , then $(x-y)(y-z)(z-x)|f(x,y,z)$ .
To prove this, just notice that $f(x,y,y)=f(x,y,x)=f(x,x,z)=0$ .

This post has been edited 5 times. Last edited by rkm0959, Aug 18, 2015, 2:31 PM

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chezbgone

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chezbgone

#23 h Aug 18, 2015, 2:39 PM • 3 Y

Y by Eugenis, Adventure10, Mango247

Alternatively?

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trumpeter

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trumpeter

#25 h Aug 18, 2015, 3:52 PM • 1 Y

Y by Adventure10

@chezbgone2 $a^2+b^2+c^2\geq ab+bc+ca$ can be derived using a multitude of other ways, including Rearrangement and Trivial, so I think this works under the restrictions.

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Swift12318

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Swift12318

#26 h Aug 18, 2015, 4:40 PM • 2 Y

Y by Adventure10, Mango247

Eugenis wrote:

Prove $\dfrac{a^3+b^3+c^3}{3}\geq abc$ without using AM-GM for $a, b, c \ge 0.$

$a,b,c\geq 0 \Rightarrow \boxed{ (a+b+c) \geq 0}$ .We also have $(a^2+b^2) \geq 2ab ; (b^2+c^2) \geq 2bc ; (a^2+c^2) \geq 2ac$ . Adding these, we get $(a^2+b^2+c^2) \geq (ab+bc+ac) \geq 0 \Rightarrow \boxed{ (a^2+b^2+c^2-ab-bc-ac) \geq 0}$ .Hence, we have $(a^3+b^3+c^3-3abc) = (a+b+c)(a^2+b^2+c^2-ab-bc-ac) \geq 0 \Rightarrow \dfrac{a^3+b^3+c^3}{3}\geq abc$ .