Question

If a,b,c ≥0,show that a(a-b)(a-c) + b(b-c)(b-a)+c(c-a)(c-b)

Answer 1

SOLUTION

TO PROVE

If a,b,c ≥0 , show that

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

EVALUATION

Case : 1

Let a = b = c

LHS = 0, since each term is 0

Case : 2

Let two of a, b, c are equal. a = b say

LHS = c(c-a)² ≥ 0 Since c ≥ 0 , (c-a)² ≥ 0

Case : 3

No two of a, b, c are equal

Without loss of generality let a > b > c

Then a - b > 0, b - c > 0, a - c > 0

Now

a(a-b)(a-c) + b(b-c)(b-a)

= ( a - b ) [ a(a-c) - b(b-c) ]

Since a > b > c

∴ a - c > b - c > 0

Also a > b

∴ a(a-b)(a-c) + b(b-c)(b-a) > 0

Also we have c(c-a)(c-b) ≥ 0

∴ a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b) ≥ 0

Combining all three cases

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

Hence proved

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Answer 2

Answer:

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