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

Question

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

Professor PC Thomas Classes · Answer

(b) : Given that $a, b, c$ are distinct +ve numbers. The expression whose sign is to be checked is\ $(b+c-a)(c+a-b)(a+b-c)-a b c$\ As this expression is symmetric in $a, b, c$ without loss of generality, we can assume that $0 G.M. for distinct real numbers)\ $\left.\begin{array}{l}\frac{x+y}{2}>\sqrt{x y} \ \frac{y+z}{2}>\sqrt{y z}\end{array} ight\} \Rightarrow\left(\frac{x+y}{2} ight)\left(\frac{y+z}{2} ight)\left(\frac{z+x}{2} ight)>x y z$\ $\frac{z+x}{2}>\sqrt{z x}$\ $\Rightarrow a b c>(a+b-c)(b+c-a)(c+a-b)$\ $\Rightarrow(b+c-a)(c+a-b)(a+b-c)-a b c<0$\ Case II : If $a+b-c=-\mathrm{ve}$ then\ $(b+c-a)(c+a-b)(a+b-c)-a b c$\ $=(+\mathrm{ve})(+\mathrm{ve})(-\mathrm{ve})-(+\mathrm{ve})$\ $=(-\mathrm{ve})-(+\mathrm{ve})=(-\mathrm{ve})$\ $\Rightarrow \quad(b+c-a)(c+a-b)(a+b-c)