
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
(ab)^3 + (bc)^3 + (ca)^3
(where dotted line shows it is a fraction)
I have no idea what to do here, help!!

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
(ab)^3 + (bc)^3 + (ca)^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 $\displaystyle x=a^2b^2,y=b^2c^2,z=c^2a^2$
in the denominator take $\displaystyle x=ab,y=bc,z=ca$.
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.

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 $\displaystyle x=a^2b^2,y=b^2c^2,z=c^2a^2$
in the denominator take $\displaystyle x=ab,y=bc,z=ca$.
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 $\displaystyle x=a^2b^2$ and then $\displaystyle x=ab$... It makes no sense!

Re: Prove positive real numbers a,b,c.
Quote:
Originally Posted by
Also sprach Zarathustra First you take $\displaystyle x=a^2b^2$ and then $\displaystyle x=ab$... 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
$\displaystyle (a^2b^2)^3+(b^2c^2)^3+(c^2a^2)^3= 3(a^2b^2)(b^2c^2)(c^2a^2)$
$\displaystyle (ab)^3+(bc)^3+(ca)^3=3(ab)(bc)(ca)$
now do you understand what i was trying to say??

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?

Re: Prove positive real numbers a,b,c.
I'm still confused, i dont get what you mean!!

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.

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.

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 $\displaystyle (a^2b^2)+(b^2c^2)+(c^2a^2)=0$ so we get $\displaystyle (a^2b^2)^3+(b^2c^2)^3+(c^2a^2)^3=3(a^2b^2)(b^2c^2)(c^2a^2)$
if you agree with this then i can continue. also tell me why you agree with this in case you do.

Re: Prove positive real numbers a,b,c.

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

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^2b^2) and y=....etc, you've just shown an expanded version of this. Hence it is true.
now look at the denominator.
we have $\displaystyle (ab)+(bc)+(ca)=0$ so we have $\displaystyle (ab)^3+(bc)^3+(ca)^3=3(ab)(bc)(ca)$.
agreed? why, why not?

Re: Prove positive real numbers a,b,c.
Same rule as before, x^3 + y^3 + z^3 = (x^2+y^2+z^2xyxzyz) +3xyz.... simplifies down to x+y+z = 3xyz. So yes, i do agree.

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^2xyxzyz) +3xyz.... simplifies down to x+y+z = 3xyz. So yes, i do agree.
so $\displaystyle [(a^2b^2)^3+(b^2c^2)^3+(c^2a^2)^3]/[(ab)^3+(bc)^3+(ca)^3]= \text{??}$ when simplified using what we have just discussed.

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^2xyxzyz) +3xyz.... simplifies down to x+y+z = 3xyz. no!! So yes, i do agree.
theres a mistake i pointed out in red.