prove that (a+b)(b+c)(c+a) =/> 8abc

chinyew

prove that (a+b)(b+c)(c+a) =/> 8abc
for all a,b,c =/> 0
any1 pls.. thx.

CompuChip

Let's start by ordering them from largest to smallest: [itex]a \ge b \ge c[/itex].
Then you can open the brackets and get 8 terms, two of which are precisely equal to abc. Picking one at random, say, b²c, can you show that this is larger than abc?

mathman

abc≥b²c, so the answer to your question is no.

thanhson95

You should use AM-GM inequality wich
[tex]\sqrt{ab}[/tex] </= (a+b)/2

Diffy

Given a,b,c =/> 0 implies
a,b,c < 0 implies
(a+b) < a
(b+c) < b
(c+a) < c
.....

Mentallic

He was meant to say that =/> is read "equal or more".

Compuchip's approach will be easiest.

Yuqing

When expanded you get

[tex]a^{2}b + a^{2}c + ab^{2} + ac^{2} + b^{2}c + bc^2 + abc + abc \geq 8abc[/tex]

If you apply GM-AM inequality to the collection you get:

[tex]a^{2}b + a^{2}c + ab^{2} + ac^{2} + b^{2}c + bc^2 + abc + abc \geq 8\sqrt[8]{a^{8}b^{8}c^{8}}[/tex]