04-09-09 Inequality..

This is the Schur's inequality.
In fact one can prove that for all real k, we have

a^k (a-b)(a-c) + b^k(b-c)(b-a) + c^k(c-a)(c-b) ≥ 0

here a, b,c are non-negative

@ Kaymant sir, Nishant sir...can you pls throw some more light?

Ohh..okk..got it :)

since the inequality is symmetrical in a , b , c
hence there is no loss in generality in assuming a\geq b\geq c
we have
a(b-a)(c-a) + b( c-b)(a-b) + c(a-c)(b-c)
= (b-a){a(c-a) - b(c-b)} + c(a-c)(b-c)
= (b-a){ca - a²-bc+b²} + c(a-c)(b-c)
= (b-a){-c(b-a)+b²-a²} + c(a-c)(b-c)
= (b-a){-c(b-a)+(b-a)(b+a)} + c(a-c)(b-c)
= (b-a)²{a+b-c} + c(a-c)(b-c)
which is non-negative becoz each term of RHS is non-negative since a\geq b\geq c and a,b,c be non-negative
thus
a(b-a)(c-a) + b( c-b)(a-b) + c(a-c)(b-c) \geq 0
hence proved [1]

asumming a≥b≥c

a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)≥0

=>(a-b) [a(a-c) -b(b-c)] +c (a-c) (b-c) ≥0(which is true)