IIT-JEE 1986Question 1
If $a$, $b$ and $c$ are distinct positive numbers, then the expression $(b+c-a)(c+a-b)(a+b-c)-abc$ is
Quadratic Equation · Miscellaneous
Solution
Concept. Each pair of factors multiplies to a square minus a square, so it is at most that square. Formulas used. (1) $(b+c-a)(c+a-b)=c^2-(a-b)^2\le c^2$, and similarly for the other pairsStep 1 - At most one factor is $\le0$. The sum of any two factors is positive (e.g. $(b+c-a)+(c+a-b)=2c>0$). If one factor is $\le0$, the product is $\le0<abc$, so the expression is negative. Step 2 - All factors positive. Multiply the three forms of (1): $$\big[(b+c-a)(c+a-b)(a+b-c)\big]^2\le a^2b^2c^2.$$ Equality would need $a=b$, $b=c$ and $c=a$, so for distinct numbers it is strict. Hence the product is $<abc$. Step 3 - Conclude. In every case $(b+c-a)(c+a-b)(a+b-c)-abc<0$. Answer: (B)