Prove positive real numbers a,b,c.

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• Jul 5th 2011, 06:11 PM
pikachu26134
Prove positive real numbers a,b,c.
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!!
• Jul 5th 2011, 07:07 PM
abhishekkgp
Re: Prove positive real numbers a,b,c.
Quote:

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 $\displaystyle x+y+z=0$ then $\displaystyle x^3+y^3+z^3=3xyz$.
In the numerator take $\displaystyle x=a^2-b^2,y=b^2-c^2,z=c^2-a^2$
in the denominator take $\displaystyle x=a-b,y=b-c,z=c-a$.
then the LHS is equal to $\displaystyle (a+b)(b+c)(c+a)$. now try to prove that for positive $\displaystyle a,b,c$, $\displaystyle (a+b)(b+c)(c+a)>8abc$. Its easy. Use AM>GM.
• Jul 6th 2011, 01:01 AM
Also sprach Zarathustra
Re: Prove positive real numbers a,b,c.
Quote:

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

First you take $\displaystyle x=a^2-b^2$ and then $\displaystyle x=a-b$... It makes no sense!
• Jul 6th 2011, 01:23 AM
abhishekkgp
Re: Prove positive real numbers a,b,c.
Quote:

Originally Posted by Also sprach Zarathustra
First you take $\displaystyle x=a^2-b^2$ and then $\displaystyle x=a-b$... It makes no sense!

In order to use the formula $\displaystyle x^3+y^3+z^3=3xyz$ if $\displaystyle x+y+z=0$ you have to appropriately set $\displaystyle x,y \, and \, z$. we get
$\displaystyle (a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3= 3(a^2-b^2)(b^2-c^2)(c^2-a^2)$
$\displaystyle (a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)$
now do you understand what i was trying to say??
• Aug 3rd 2011, 10:48 PM
pikachu26134
Re: Prove positive real numbers a,b,c.
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?
• Aug 6th 2011, 11:01 PM
pikachu26134
Re: Prove positive real numbers a,b,c.
I'm still confused, i dont get what you mean!!
• Aug 6th 2011, 11:06 PM
abhishekkgp
Re: Prove positive real numbers a,b,c.
Quote:

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 $\displaystyle x+y+z=0$ then $\displaystyle x^3+y^3+z^3=3xyz$. 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.
• Aug 6th 2011, 11:12 PM
pikachu26134
Re: Prove positive real numbers a,b,c.
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.
• Aug 6th 2011, 11:20 PM
abhishekkgp
Re: Prove positive real numbers a,b,c.
Quote:

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 $\displaystyle (a^2-b^2)+(b^2-c^2)+(c^2-a^2)=0$ so we get $\displaystyle (a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3=3(a^2-b^2)(b^2-c^2)(c^2-a^2)$

if you agree with this then i can continue. also tell me why you agree with this in case you do.
• Aug 6th 2011, 11:26 PM
abhishekkgp
Re: Prove positive real numbers a,b,c.
• Aug 6th 2011, 11:27 PM
pikachu26134
Re: Prove positive real numbers a,b,c.
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.
• Aug 6th 2011, 11:29 PM
abhishekkgp
Re: Prove positive real numbers a,b,c.
Quote:

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 $\displaystyle (a-b)+(b-c)+(c-a)=0$ so we have $\displaystyle (a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)$.
agreed? why, why not?
• Aug 6th 2011, 11:32 PM
pikachu26134
Re: Prove positive real numbers a,b,c.
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.
• Aug 6th 2011, 11:40 PM
abhishekkgp
Re: Prove positive real numbers a,b,c.
Quote:

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 $\displaystyle [(a^2-b^2)^3+(b^2-c^2)^3+(c^2-a^2)^3]/[(a-b)^3+(b-c)^3+(c-a)^3]= \text{??}$ when simplified using what we have just discussed.
• Aug 6th 2011, 11:42 PM
abhishekkgp
Re: Prove positive real numbers a,b,c.
Quote:

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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