Question

If $(a,b,c)\in +R$ such that $b+c-a,c+a-b,a+b-c$ are positive, then the expression
$(b+c-a)(a+b-c)(c+a-b)-abc$ is

• A. Positive
• B. Negative
• C. Zero
• D. None of the above.

Answer 1

The correct option is B Negative

we know that $A.M.\ge G.M.$

$⟹\frac{x+y}{2}\ge \sqrt{xy}$

let $x=b+c-a$ and $y=a+b-c$

therefore, $\frac{b+c-a+a+b-c}{2}\ge \sqrt{(b+c-a)(a+b-c)}$

$⟹\frac{2b}{2}\ge \sqrt{(b+c-a)(a+b-c)}$

$⟹b\ge \sqrt{(b+c-a)(a+b-c)}$ ----------(1)

similarly, let $x=b+c-a$ and $y=c+a-b$

therefore, $\frac{b+c-a+c+a-b}{2}\ge \sqrt{(b+c-a)(c+a-b)}$

$⟹\frac{2c}{2}\ge \sqrt{(b+c-a)(c+a-b)}$

$⟹c\ge \sqrt{(b+c-a)(c+a-b)}$ ----------(2)

similarly, let $x=a+b-c$ and $y=c+a-b$

therefore, $\frac{a+b-c+c+a-b}{2}\ge \sqrt{(a+b-c)(c+a-b)}$

$⟹\frac{2a}{2}\ge \sqrt{(a+b-c)(c+a-b)}$

$⟹a\ge \sqrt{(a+b-c)(c+a-b)}$ ----------(3)

multiplying (1), (2), (3) we get

$abc\ge \sqrt{(b+c-a{)}^2(a+b-c{)}^2(c+a-b{)}^2}$

$⟹abc\ge (b+c-a)(a+b-c)(c+a-b)$

$⟹(b+c-a)(a+b-c)(c+a-b)\le abc$

$⟹(b+c-a)(a+b-c)(c+a-b)-abc\le 0$

therefore, $(b+c-a)(a+b-c)(c+a-b)-abc$ is negative