Thread: Proofing

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

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......?

Originally Posted by sa-ri-ga-ma

Minimum positive real numbers for a, b and c will be 1.

What if ?

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??

OK.
Now expand the given expression, and divide it by abc. You get

= 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.