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

prove that (a+b)(b+c)(c+a) =/> 8abc
for all a,b,c =/> 0
any1 pls.. thx.
 Blog Entries: 5 Recognitions: Homework Help Science Advisor 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?

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

You should use AM-GM inequality wich
$$\sqrt{ab}$$ </= (a+b)/2
 Given a,b,c =/> 0 implies a,b,c < 0 implies (a+b) < a (b+c) < b (c+a) < c .....
 Recognitions: Homework Help He was meant to say that =/> is read "equal or more". Compuchip's approach will be easiest.
 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: $$a^{2}b + a^{2}c + ab^{2} + ac^{2} + b^{2}c + bc^2 + abc + abc \geq 8\sqrt[8]{a^{8}b^{8}c^{8}}$$