1. ## Proofing

If a, b and c are positive real numbers, prove that:
bc(b+c)+ca(c+a)+ab(a+b)≥6abc

2. Originally Posted by cedricc
If a, b and c are positive real numbers, prove that:
bc(b+c)+ca(c+a)+ab(a+b)≥6abc
Minimum positive real numbers for a, b and c will be 1.
So the expression bc(b+c)+ca(c+a)+ab(a+b) becomes......?
For any other values for a, b and c more one, the given expression will be......?

3. Originally Posted by sa-ri-ga-ma
Minimum positive real numbers for a, b and c will be 1.
What if $a = 0 + \epsilon$ ?

4. Originally Posted by sa-ri-ga-ma
Minimum positive real numbers for a, b and c will be 1.
So the expression bc(b+c)+ca(c+a)+ab(a+b) becomes......?
For any other values for a, b and c more one, the given expression will be......?
but how do i reason this??

5. OK.
Now expand the given expression, and divide it by abc. You get
$[b^2c + bc^2 + c^2a +ca^2 + a^2b + ab^2]/abc$
= b/a + c/a + c/b + a/b + a/c + b/c
Now consider
b/a + a/b = (b^2 + a^2)/ab = [(a+b)^2- 2ab]/ab = [(a+b)^2/ab] -2 > 0
because [(a+b)^2/ab] is always greater than 2.
Similarly you can proceed for other terms in the expressions.