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: $a \ge b \ge c$.
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 which
$$\sqrt{ab}$$ </= (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

$$a^{2}b + a^{2}c + ab^{2} + ac^{2} + b^{2}c + bc^2 + abc + abc \geq 8abc$$

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

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