Math Problem Statement

Given any two vectors in \( \mathbb{R}^n \): \(\vec{a} = (a_1, a_2, \dots, a_n)\), \(\vec{b} = (b_1, b_2, \dots, b_n)\), show that \(\vec{a} \cdot \vec{b} \leq \frac{1}{\epsilon} |\vec{a}|^2 + \frac{1}{4\epsilon} |\vec{b}|^2\) for any \(\epsilon > 0\). (30 pts) [You may assume that the Cauchy-Schwarz inequality holds.]

Solution

The question asks to prove that for any two vectors a\vec{a} and b\vec{b} in Rn\mathbb{R}^n, and for any ϵ>0\epsilon > 0, the following inequality holds:

ab1ϵa2+14ϵb2\vec{a} \cdot \vec{b} \leq \frac{1}{\epsilon} |\vec{a}|^2 + \frac{1}{4\epsilon} |\vec{b}|^2

Additionally, we are allowed to use the Cauchy-Schwarz inequality in the proof.

Step-by-Step Solution:

  1. Recall the Cauchy-Schwarz inequality:
    The Cauchy-Schwarz inequality in Rn\mathbb{R}^n states that for any vectors a\vec{a} and b\vec{b}: abab|\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}| where a|\vec{a}| and b|\vec{b}| represent the Euclidean norms (or lengths) of the vectors a\vec{a} and b\vec{b}.

  2. Start with the expression ab\vec{a} \cdot \vec{b}: By applying the inequality ab12(ϵa2+1ϵb2)ab \leq \frac{1}{2} \left( \epsilon a^2 + \frac{1}{\epsilon} b^2 \right), where a=aa = |\vec{a}| and b=bb = |\vec{b}|, we proceed to prove the required inequality.

  3. Apply the inequality ab12(ϵa2+1ϵb2)ab \leq \frac{1}{2} \left( \epsilon a^2 + \frac{1}{\epsilon} b^2 \right):
    Let a=aa = |\vec{a}| and b=bb = |\vec{b}|. This gives us: abab\vec{a} \cdot \vec{b} \leq |\vec{a}| |\vec{b}| Now, using the inequality ab12(ϵa2+1ϵb2)ab \leq \frac{1}{2} \left( \epsilon a^2 + \frac{1}{\epsilon} b^2 \right), we substitute: ab12(ϵa2+1ϵb2)|\vec{a}| |\vec{b}| \leq \frac{1}{2} \left( \epsilon |\vec{a}|^2 + \frac{1}{\epsilon} |\vec{b}|^2 \right)

  4. Manipulate to match the required inequality: Multiply both sides by 2 to eliminate the 12\frac{1}{2}: 2abϵa2+1ϵb22 |\vec{a}| |\vec{b}| \leq \epsilon |\vec{a}|^2 + \frac{1}{\epsilon} |\vec{b}|^2 Now divide both sides by 2ϵ2 \epsilon to get the desired inequality: ab1ϵa2+14ϵb2\vec{a} \cdot \vec{b} \leq \frac{1}{\epsilon} |\vec{a}|^2 + \frac{1}{4\epsilon} |\vec{b}|^2

Thus, the inequality is proven using the Cauchy-Schwarz inequality.

Would you like further details on this proof or have any questions? Here are five related questions to expand your understanding:

  1. What are the conditions under which the Cauchy-Schwarz inequality becomes an equality?
  2. How would this inequality change if ϵ\epsilon were negative?
  3. Can you apply a similar technique to prove other inequalities involving vector norms?
  4. How does this inequality relate to the general method of bounding inner products?
  5. How can this inequality be used in optimization problems involving vectors?

Tip: The technique of bounding products using inequalities like Cauchy-Schwarz and manipulating expressions with ϵ\epsilon is often useful when proving inequalities in both linear algebra and analysis.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Inequalities

Formulas

Cauchy-Schwarz Inequality
General inequality involving products and sums

Theorems

Cauchy-Schwarz Inequality

Suitable Grade Level

Undergraduate Level