1Ohh..okk..got it :)
Show that a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)
cannot be negative
This i saw in a website online.. i liked this one because it does not even require AM GM ;)
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5 Answers
66This is the Schur's inequality.
In fact one can prove that for all real k, we have
ak (a-b)(a-c) + bk(b-c)(b-a) + ck(c-a)(c-b) ≥ 0
here a, b,c are non-negative
1since 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]
1asumming 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)