2014/013) If a,b,c are positive numbers, show that 8(a^3+b^3+c^3)≥(a+b)^3+(a+c)^3+(b+c)^3

we have a3+b^3–a^2b–b^2a 
= a3−a^2b–b^2a+b^3
= a2(a−b)−b^2(a−b)=(a^2−b^2)(a−b)=(a+b)(a−b)^2>=0

Hence 
a3+b^3>=a^2b+b^2a

Multiply by 3 and add a3+b^3 on both sides

4(a^3+b^3)>=a^3+b^3+3(a^2b+b^2a)>=(a+b)^3
4(a^3+b^3)>=(a+b) .. (1)

Similarly

4(b^3+c^3)>=(b+c) ... (2)
4(c^3+a^3)>=(c+a) ...(3)

Adding (1), (2), (3) we get the result