Sequences & Series
Sequence and Series
Allen Star Batch
Grade 11
Question:
If $a, b, c$ are three positive real numbers, then $\frac{a^2+1}{b+c} + \frac{b^2+1}{c+a} + \frac{c^2+1}{a+b}$ can be never be equal to:
$1$
$2$
$\frac{8}{3}$
$3$
Step-by-Step Solution
Key Concept: Decompose the numerators strategically and apply Nesbitt's inequality to establish a lower bound involving symmetric expressions.
Starting with $\frac{a^2+1}{b+c} + \frac{b^2+1}{c+a} + \frac{c^2+1}{a+b} \geq 2\left(\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}\right)$, we apply the constraint that by Nesbitt's inequality the sum of reciprocals $\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b} \geq \frac{3}{2(a+b+c)}$ when expanded appropriately. The key insight shows $(a+b+c)\left(\frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{a+b}\right) \geq \frac{9}{2}$, leading to the conclusion that equality holds when $a = b = c$.
Correct Answer: 1,2,3