Question

If $a,b,c$ are distinct positive numbers, then the expression $(b+c-a)(c+a-b)(a+b-c)-abc$ is

• A. Zero
• B. Imaginary
• C. Non-positive
• D. Non-negative

Answer 1

Answer: (C)

Non-positive

Without loss of generality we can assume that $a\ge b\ge c$

Let

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

$\Rightarrow S=(-a+b+c)[a-(b-c\left)\right][a+(b-c\left)\right]$

$\Rightarrow S=(-a+b+c)[a^2-(b-c{)}^2]$

$\Rightarrow S=(-a+b+c)[a^2-b^2-c^2+2bc]$

$\Rightarrow S=-(a^3+b^3+c^3)-2abc+b^{2}c+b{c}^2+a{b}^2+a^{2}b+a{c}^2+a^{2}c$

$\Rightarrow abc-S=(a^3+b^3+c^3)+3abc-(b^{2}c+b{c}^2+a{b}^2+a^{2}b+a{c}^2+a^{2}c)$

$\Rightarrow abc-S=(a^3-a^{2}b)+(b^3-b^{2}c)+(c^3-c^{2}a)+(abc-b{c}^2)+(abc-a{b}^2)+(abc-a^{2}c)$

$\Rightarrow abc-S=a^2(a-b)+b^2(b-c)+c^2(c-a)+bc(a-c)+ab(c-b)+ac(b-a)$

$\Rightarrow abc-S=a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)$

$\Rightarrow abc-S=(a-b)\left[a\right(a-c)-b(b-c\left)\right]+c(c-a)(c-b)$

$\Rightarrow abc-S=(a-b)2[a^2-b^2+c(b-a\left)\right]+c(c-a)(c-b)$

$\Rightarrow abc-S=(a-b{)}^2[a+b-c]+c(c-a\left)\right(c-b)$

Now $(c-a)\le 0$ and $(c-b)\le 0$

$\Rightarrow c(c-a)(c-b)\ge 0$

and

$(a-b{)}^2(a+b-c)\ge 0$

This shows

$abc-S\ge 0$

$\Rightarrow S-abc\le 0\Rightarrow (b+c-a)(c+a-b)(a+b-c)-abc\le 0$

Hence, the expression is non-positive.

So, $\left(C\right)$