A,b,c are three distinct positive numbers. Prove that ba^2+ab^2+bc^2+cb^2+ca^2+ac^2<2(a^3+b^3+c^3) ?

1 Answer
Nov 1, 2017

See below.

Explanation:

Considering

f(a,b,c) =2(a^3+b^3+c^3)-( ba^2+ab^2+bc^2+cb^2+ca^2+ac^2)

as we can verify f(a,b,c) is symmetric regarding the order of it's arguments. We should prove that

f(a,b,c) ge 0 for a > 0,b > 0,c > 0

Due to symmetry, min f(a,b,c) = min f(x_0,x_0,x_0) = 0

Supposing now epsilon such that x_0+epsilon > 0

f(x_0,x_0,x_0+epsilon) = 2epsilon^2(2x_0+epsilon) > 0 so

f(a,b,c) has minimum 0 value for a=b=c and is positive for any other choice of a > 0, b > 0, c > 0 so this demonstrates that

2(a^3+b^3+c^3) ge ( ba^2+ab^2+bc^2+cb^2+ca^2+ac^2)