Question

Show that the expressions
$a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)$
and $a^2(a-b)(a-c)+b^2(b-c)(b-a)+c^2(c-a)(c-b)$
are both positive.

Answer 1

$S=a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)$

$S=(a-b)(a(a-c)-b(b-c))+c(c-a)(c-b)$

As $a>b>c$

$a-b>0$ and $a-c>b-c$

Thus , $a(a-c)>b(b-c)$

$a^2(a-c)>b^2(b-c)$

$a(a-c)-b(b-c)>0$ and $(c-a)(c-b)>0$ because both are negative therfore multiplication of two neagtive is positive

Therfore ,

$S=a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)>0$

$I=a^2(a-b)(a-c)+b^2(b-c)(b-a)+c^2(c-a)(c-b)$

$I=(a-b)(a^2(a-c)-b^2(b-c))+c^2(c-a)(c-b)$

$I>0$ as , $a^2(a-c)-b^2(b-c)>0$ and $(c-a)(c-b)>0$ because both are negative therfore multiplication of two neagtive is positive