Sequences & Series
Sequence and Series
star_batch_jee_advanced_2025
Grade 11
Question:
Let $a, b, c$ be positive real number such that $a + b + c \geq 4$, then find the minimum value of $\frac{a^3}{(a-b)(a-c)} + \frac{b^3}{(b-c)(b-a)} + \frac{c^3}{(c-a)(c-b)}$.
Step-by-Step Solution
Key Concept: Factoring the numerator as a cyclic sum and recognizing the common denominator structure reveals that the expression simplifies to $a + b + c$ under appropriate constraints.
The expression $\frac{a^3}{(a-b)(c-a)} + \frac{b^3}{(b-c)(a-b)} + \frac{c^3}{(c-a)(b-c)}$ is rewritten using algebraic manipulation. The numerator becomes $a^3(b-c) + b^3(c-a) + c^3(a-b)$ divided by the product of differences. Through factorization and simplification, this equals $-(a+b+c) - \frac{(b-c)(c-a)(a-b)(-a-b-c)}{(a-b)(b-c)(c-a)} = a + b + c \geq 4$.
Correct Answer: 4