Prove positive real numbers a,b,c.

**pikachu26134** · July 5th 2011, 06:11 PM

Prove that for any three distinct positive real numbers a, b and c:
((a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3
----------------------------------------------------------- > 8abc
(a-b)^3 + (b-c)^3 + (c-a)^3

(where dotted line shows it is a fraction)
I have no idea what to do here, help!!

**abhishekkgp** · July 5th 2011, 07:07 PM

Originally Posted by pikachu26134

Prove that for any three distinct positive real numbers a, b and c:
((a^2 - b^2)^3 + (b^2 - c^2)^3 + (c^2 - a^2)^3
----------------------------------------------------------- > 8abc
(a-b)^3 + (b-c)^3 + (c-a)^3

(where dotted line shows it is a fraction)
I have no idea what to do here, help!!

there's an interesting property that if $x+y+z=0$ then $x^3+y^3+z^3=3xyz$ .
In the numerator take $x=a^2-b^2,y=b^2-c^2,z=c^2-a^2$
in the denominator take $x=a-b,y=b-c,z=c-a$ .
then the LHS is equal to $(a+b)(b+c)(c+a)$ . now try to prove that for positive $a,b,c$ , $(a+b)(b+c)(c+a)>8abc$ . Its easy. Use AM>GM.

**Also sprach Zarathustra** · July 6th 2011, 01:01 AM

Originally Posted by abhishekkgp

there's an interesting property that if $x+y+z=0$ then $x^3+y^3+z^3=3xyz$ .
In the numerator take $x=a^2-b^2,y=b^2-c^2,z=c^2-a^2$
in the denominator take $x=a-b,y=b-c,z=c-a$ .
then the LHS is equal to $(a+b)(b+c)(c+a)$ . now try to prove that for positive $a,b,c$ , $(a+b)(b+c)(c+a)>8abc$ . Its easy. Use AM>GM.

First you take $x=a^2-b^2$ and then $x=a-b$ ... It makes no sense!

**abhishekkgp** · July 6th 2011, 01:23 AM

Originally Posted by Also sprach Zarathustra

First you take $x=a^2-b^2$ and then $x=a-b$ ... It makes no sense!

In order to use the formula $x^3+y^3+z^3=3xyz$ if $x+y+z=0$ you have to appropriately set $x,y \, and \, z$ . we get
$(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3= 3(a^2-b^2)(b^2-c^2)(c^2-a^2)$
$(a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)$
now do you understand what i was trying to say??

**pikachu26134** · August 3rd 2011, 10:48 PM

Im a little confused, are you trying to say that (a+b)/2 (b+c)/2 (c+a)/2 >(or equal) to square root ab * square root bc *square root ca?

**pikachu26134** · August 6th 2011, 11:01 PM

I'm still confused, i dont get what you mean!!

**abhishekkgp** · August 6th 2011, 11:06 PM

Originally Posted by pikachu26134

I'm still confused, i dont get what you mean!!

okay then. for the time being forget about the question you have posted. Try to prove that if $x+y+z=0$ then $x^3+y^3+z^3=3xyz$ . Post the proof of this in your reply and i will tell you next step. This way you will understand what i was trying to say.

**pikachu26134** · August 6th 2011, 11:12 PM

Wait, first of all i dont know how to prove that. But, wouldn't (coming back to my question) it end up (a+b)/2 x (b+c)/2 x (c+a)/2 = sqr ab x sqr bc x sqr ca ?? (using AM>GM inequality) I need to get this done by tomorrow so yeah.

**abhishekkgp** · August 6th 2011, 11:20 PM

Originally Posted by pikachu26134

Wait, first of all i dont know how to prove that. But, wouldn't (coming back to my question) it end up (a+b)/2 x (b+c)/2 x (c+a)/2 = sqr ab x sqr bc x sqr ca ?? (using AM>GM inequality) I need to get this done by tomorrow so yeah.

it can be proved by cubing both sides or by theory of equations. Suppose its been proved.
then look at the numerator. we have $(a^2-b^2)+(b^2-c^2)+(c^2-a^2)=0$ so we get $(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3=3(a^2-b^2)(b^2-c^2)(c^2-a^2)$

if you agree with this then i can continue. also tell me why you agree with this in case you do.

**abhishekkgp** · August 6th 2011, 11:26 PM

If x+y+z=0,show that x3+y3+z3=3xyz? - Yahoo! Answers India
read this too.

**pikachu26134** · August 6th 2011, 11:27 PM

Yes i do agree with this, because you simply cubed both sides, and as x=(a^2-b^2) and y=....etc, you've just shown an expanded version of this. Hence it is true.

**abhishekkgp** · August 6th 2011, 11:29 PM

Originally Posted by pikachu26134

Yes i do agree with this, because you simply cubed both sides, and as x=(a^2-b^2) and y=....etc, you've just shown an expanded version of this. Hence it is true.

now look at the denominator.
we have $(a-b)+(b-c)+(c-a)=0$ so we have $(a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)$ .
agreed? why, why not?

**pikachu26134** · August 6th 2011, 11:32 PM

Same rule as before, x^3 + y^3 + z^3 = (x^2+y^2+z^2-xy-xz-yz) +3xyz.... simplifies down to x+y+z = 3xyz. So yes, i do agree.

**abhishekkgp** · August 6th 2011, 11:40 PM

Originally Posted by pikachu26134

Same rule as before, x^3 + y^3 + z^3 = (x^2+y^2+z^2-xy-xz-yz) +3xyz.... simplifies down to x+y+z = 3xyz. So yes, i do agree.

so $[(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3]/[(a-b)^3+(b-c)^3+(c-a)^3]= \text{??}$ when simplified using what we have just discussed.

**abhishekkgp** · August 6th 2011, 11:42 PM

Originally Posted by pikachu26134

Same rule as before, x^3 + y^3 + z^3 = (x^2+y^2+z^2-xy-xz-yz) +3xyz.... simplifies down to x+y+z = 3xyz. no!! So yes, i do agree.

theres a mistake i pointed out in red.

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