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)# ?

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Answer 1 · 2017-11-01T14:49:16

See below.

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)#