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

Well it would equal 3xyz/(a-b)^3 + (b-c)^3 + (c-a)^3 wouldnt it? I dont know how to simplify the denomiator.

Originally Posted by pikachu26134

Well it would equal 3xyz/(a-b)^3 + (b-c)^3 + (c-a)^3 wouldnt it? I dont know how to simplify the denomiator.

see post #9 for numerator and post#12 for denominator. then post the answer again. don't use x,y and z in the answer. use only a, b and c.

[3(a^2-b^2)(b^2-c^2)(c^2-a^2)]/ [3(a-b)(b-c)(c-a)] is that right?

Originally Posted by pikachu26134

[3(a^2-b^2)(b^2-c^2)(c^2-a^2)]/ [3(a-b)(b-c)(c-a)] is that right?

perfect. now can you further simplify this. use $\displaystyle (a^2-b^2)=(a-b)(a+b)$. etc. what do you get?

You then get (a+b)(b+c)(c+a) > 8abc

Originally Posted by pikachu26134

You then get (a+b)(b+c)(c+a) > 8abc

can you please post that here in more detail.

[(a+b)(a-b)(b+c)(b-c)(c-a)(c+a)]/[(a-b)(b-c)(c-a)] and then you simply it to [(a+b)(b+c)(c+a)]/1 = (a+b)(b+c)(c+a)

Originally Posted by pikachu26134

[(a+b)(a-b)(b+c)(b-c)(c-a)(c+a)]/[(a-b)(b-c)(c-a)] and then you simply it to [(a+b)(b+c)(c+a)]/1 = (a+b)(b+c)(c+a)

that's perfect!
now can i have some rep points.

Yup, ive thanked on all your posts. What do I do from there? I presume its the quadratic rule where either (a+b)>8abc etc.... but what exactly do i do to find sets abc?

Originally Posted by pikachu26134

Yup, ive thanked on all your posts. What do I do from there? I presume its the quadratic rule where either (a+b)>8abc etc.... but what exactly do i do to find sets abc?

$\displaystyle (\sqrt{a}- \sqrt{b})^2>0$
$\displaystyle \Rightarrow a+b-2 \sqrt{ab}>0$

$\displaystyle \Rightarrow a+b>2 \sqrt{ab}$ .......(1)

similarly $\displaystyle b+c> 2 \sqrt{bc}$ ......(2)

and $\displaystyle c+a > 2 \sqrt{ca}$ ...... (3)
multiply the above three inequalities.

2abc+a^2b+ac^2+a^2c+b^2c+ab^2+bc^2 > 2sqr ab x 2sqr bc x 2sqr ca

Originally Posted by pikachu26134

2abc+a^2b+ac^2+a^2c+b^2c+ab^2+bc^2 > 2sqr ab x 2sqr bc x 2sqr ca

well what i was expecting was
$\displaystyle (a+b)(b+c)(c+a)> 2 \sqrt{ab} \cdot 2 \sqrt{bc} \cdot 2 \sqrt{ca}=8 \sqrt{abbcca}=8abc$
so $\displaystyle (a+b)(b+c)(c+a)>8abc$

Oh right sorry, I thought you meant expanding the whole lot. Yes, that is correct. But now what do I do? I realise you simplified the RHS to get 8abc.

Originally Posted by pikachu26134

Oh right sorry, I thought you meant expanding the whole lot. Yes, that is correct. But now what do I do? I realise you simplified the RHS to get 8abc.

what is left? which part do you think remains unproven?

It remains to find groups abc such that that equation is true