Question

If a, b, c be non-negative numbers, show that :

a²(a − b)(a − c) + b²(b − c)(b − a) + c²(c − a)(c − b) ≥ 0.

Answer 1

Since the give inequality is symmetrical in a, b,c there is no loss of generality in assuming that

a ≥ b ≥ c.

∴ a² (a − b)(a − c) + b² (b − c)(b − a) + c² (c − a)(c − b)

= (a − b) {a² (a − c) − b² (b − c)} + c² (c − a)(c − b)

= (a − b) {(a³ − b³ ) − c(a² − b² )} + c² (a − c)(b − c)

= (a − b) {(a − b)(a² + ab + b² ) − c (a − b)(a + b)} + c² (a − c)(b − c)

= (a − b)² {(a² + ab + b² − c(a + b)} + c² (a − c)(b − c)}

= (a − b)² {(a² + ab + b² − ca − c} + c² (a − c)(b − c)

= (a − b)² {a(a − c) + b(b − c) + ab} + c² (a − c)(b − c)

Which is non-negative because each term of RHS is non-negative since a ≥ b ≥ c And a, b, c be non-negative

Thus, 

a² (a − b)(a − c) + b² (b − c)(b − a) + c² (c − a)(c − b) ≥ 0.