Question

If \( a_1, a_2, \ldots, a_n \) are any real numbers and n is any positive integer, then

• A. \( n \sum_{i=1}^n a_i^2 < (\sum_{i=1}^n a_i)^2 \)
• B. \( n \sum_{i=1}^n a_i^2 \geq (\sum_{i=1}^n a_i)^2 \)
• C. \( \sum_{i=1}^n a_i^2 \geq (\sum_{i=1}^n a_i)^2 \)
• D. None of the above

Hint:

This inequality is a direct application of the Cauchy-Schwarz inequality. It can also be seen as a relationship between the arithmetic mean of the squares and the square of the arithmetic mean: \( \frac{\sum a_i^2}{n} \geq \left(\frac{\sum a_i}{n}\right)^2 \), which simplifies to the given expression. Equality holds if and only if all \( a_i \) are equal.

Answer 1

The Correct Option is B

This question relates to the Cauchy-Schwarz inequality. The inequality in its vector form states that for two vectors \( \vec{u} \) and \( \vec{v} \), \( (\vec{u} \cdot \vec{v})^2 \leq ||\vec{u}||^2 ||\vec{v}||^2 \). Let's define two vectors in \( \mathbb{R}^n \): \[ \vec{u} = (1, 1, 1, \ldots, 1) \] \[ \vec{v} = (a_1, a_2, a_3, \ldots, a_n) \] Now, let's calculate the components of the Cauchy-Schwarz inequality for these vectors. \[\begin{array}{rl} \bullet & \text{Dot product: \( \vec{u} \cdot \vec{v} = 1 \cdot a_1 + 1 \cdot a_2 + \ldots + 1 \cdot a_n = \sum_{i=1}^n a_i \)} \\ \bullet & \text{Squared norm of \( \vec{u} \): \( ||\vec{u}||^2 = 1^2 + 1^2 + \ldots + 1^2 = n \)} \\ \bullet & \text{Squared norm of \( \vec{v} \): \( ||\vec{v}||^2 = a_1^2 + a_2^2 + \ldots + a_n^2 = \sum_{i=1}^n a_i^2 \)} \\ \end{array}\] Substitute these into the inequality \( (\vec{u} \cdot \vec{v})^2 \leq ||\vec{u}||^2 ||\vec{v}||^2 \): \[ \left(\sum_{i=1}^n a_i\right)^2 \leq (n) \left(\sum_{i=1}^n a_i^2\right) \] Rearranging this gives: \[ n \sum_{i=1}^n a_i^2 \geq \left(\sum_{i=1}^n a_i\right)^2 \] This matches option (B).