Sequences & Series
Sequence and Series
Allen Star Batch
Grade 11
Question:
Let $x_1, x_2, ..., x_{2018}$ be positive real numbers such that $x_1 + x_2 + ... + x_{2018} = 1$. Determine the smallest constant $k$ such that $k \sum_{i=1}^{2018} \frac{x_i^2}{1-x_i} \geq 1$
Step-by-Step Solution
Key Concept: Cauchy-Schwarz inequality $\frac{(1-x_1) + (1-x_2) + \cdots + (1-x_{2018})}{\sum \frac{1}{1-x_r}} \geq 2018$ yields the lower bound.
Starting with $\sum_{r=1}^{2018} \frac{x_r^2}{1-x_r} = \sum_{r=1}^{2018} [-(x_r+1) + \frac{1}{1-x_r}] = -2019 + \sum_{r=1}^{2018} \frac{1}{1-x_r}$, we apply Cauchy-Schwarz inequality to get $\sum_{r=1}^{2018} \frac{1}{1-x_r} \geq \frac{(2018)^2}{2017}$. This gives the minimum value of $k$ as $2017$ when the least value of the sum equals $2017$.
Correct Answer: 2017