Animated Solution for Mathematics - Sequence and Series: If $a,b$ and $c$ are distinct positive numbers, then the expression $(b+c-a)(c+a-b)(a+b-c)-abc$ is

The expression provided is $(b+c-a)(c+a-b)(a+b-c)-abc$. At first glance, this appears to be a complex algebraic knot, but it possesses a deep, underlying symmetry.

To simplify the expression, we employ the Substitution Bridge. We define the following variables: $x=b+c-a$ $y=c+a-b$ $z=a+b-c$

This transformation simplifies the expression to $xyz-abc$. To relate these back to the original variables, we solve for $a$, $b$, and $c$:

We must consider the signs of $x$, $y$, and $z$. If two variables were negative, their sum would be negative, implying that $a$, $b$, and $c$ are negative, which contradicts the standard premise of such problems.

Therefore, at most one variable can be negative. If one variable is negative, the product $xyz$ becomes negative, and since $abc$ is positive, the entire expression $xyz-abc$ is clearly negative.

When $x,y,z>0$, we utilize the Arithmetic Mean-Geometric Mean (AM-GM) Inequality. Since $a=\frac{y+z}{2}$, the inequality dictates:

By symmetry, we can establish the same relationship for the other variables:

Multiplying these three inequalities together yields:

This simplifies to $abc>xyz$. Consequently, it follows that:

Through the application of symmetry, substitution, and the power of inequalities, we have demonstrated that the expression is always negative. When encountering symmetric expressions in the JEE Advanced exam, look for these substitutions to simplify the structure and let the inequalities perform the heavy lifting.