If a, b and c are positive real numbers, prove that:
bc(b+c)+ca(c+a)+ab(a+b)≥6abc
OK.
Now expand the given expression, and divide it by abc. You get
$\displaystyle [b^2c + bc^2 + c^2a +ca^2 + a^2b + ab^2]/abc$
= 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.