prove the inequality:a,b,c>=0, I=ab(a+b-2c)+bc(b+c-2a)+ca(c+a-2b)>=0?

Answer 1

ab(a + b - 2c) + bc(b + c - 2a) + ca(c + a - 2b) >=? 0

a²b + ab² + b²c + bc² + ac² + a²c - 6abc >=? 0

abc(a/c + b/c + b/a + c/a + c/b + a/b) - 6abc >=? 0

abc(a/c + b/c + b/a + c/a + c/b + a/b - 6) >=? 0

Now look at your terms inside. You have:
a/c + c/a
b/c + c/b
a/b + b/a

These are reciprocals. Regardless of what numbers you pick for a, b, c (which must be 1 or higher), the sum will be no smaller than 2. (1/1 + 1/1 = 2).

At a minimum you have 6, but it can never result in something smaller than 6, therefore the amount in the parentheses will be 0 or greater. The product of this with a positive amount (abc) will also be 0 or greater.

Alternatively,
a²b + ab² + b²c + bc² + ac² + a²c - 6abc >=? 0
a(b² + c² - 2bc) + b(a² + c² -2ac) + c(a² + b² - 2ab) >=? 0
a(b-c)² + b(a-c)² + c(a-b)² >=? 0

Here you have positive integers and squares (which are always 0 or greater), so the sum will be greater or equal to zero.

Answer 2

expanding
a^2b+ab^2-2abc++b^2c+bc^2
-2abc+c^2a+ca^2-2abc
=a^2b+bc^2-2abc+ab^2+ac^2
-2abc+b^2c+ca^2-2abc
=b(a^2+c^2-2ac)+a(b^2+c^2-2bc)
+c(a^2+b^2-2ab)
=b(a-c)^2+a(b-c)^2+c(a-b)^2
since a,b,c>0 and the squares are always positive
the given expression is >0

Answer 3

I dont see any inequality ?