To solve this problem involving the Cauchy-Schwarz Inequality in $\mathbb{R}^2$, let's go through each step:

Step-by-Step Solution

1. Restate the Goal: The inequality $|u \cdot v| \leq \|u\|\|v\|$ is equivalent to proving $|u \cdot v|^2 \leq \|u\|^2 \|v\|^2$.

2. Expanding the LHS $|u \cdot v|^2$:

   • Given $u = (u_1, u_2)$ and $v = (v_1, v_2)$, we have: $u \cdot v = u_1 v_1 + u_2 v_2$
   • Then $|u \cdot v|^2 = (u_1 v_1 + u_2 v_2)^2$.
   • Expanding $(u_1 v_1 + u_2 v_2)^2$ gives: $(u_1 v_1 + u_2 v_2)^2 = (u_1 v_1)^2 + (u_2 v_2)^2 + 2 u_1 v_1 u_2 v_2$
   • Fill in the blank in step 2 with: $(u_1 v_1 + u_2 v_2)^2 = (u_1 v_1)^2 + (u_2 v_2)^2 + 2 u_1 v_1 u_2 v_2$

3. Expanding the RHS $\|u\|^2 \|v\|^2$:

   • We calculate $\|u\|^2 = u_1^2 + u_2^2$ and $\|v\|^2 = v_1^2 + v_2^2$.
   • Thus, $\|u\|^2 \|v\|^2 = (u_1^2 + u_2^2)(v_1^2 + v_2^2)$.
   • Expanding this product: $(u_1^2 + u_2^2)(v_1^2 + v_2^2) = u_1^2 v_1^2 + u_1^2 v_2^2 + u_2^2 v_1^2 + u_2^2 v_2^2$
   • Fill in the blank in step 3 with: $\|u\|^2 \|v\|^2 = u_1^2 v_1^2 + u_1^2 v_2^2 + u_2^2 v_1^2 + u_2^2 v_2^2$

4. Subtracting LHS from RHS:

   • We take the expanded RHS and subtract the expanded LHS: $\left( u_1^2 v_1^2 + u_1^2 v_2^2 + u_2^2 v_1^2 + u_2^2 v_2^2 \right) - \left( (u_1 v_1)^2 + (u_2 v_2)^2 + 2 u_1 v_1 u_2 v_2 \right)$
   • Simplify this expression: $= (u_1 v_2 - u_2 v_1)^2$
   • Fill in the blank in step 4 with: $(u_1 v_2 - u_2 v_1)^2$

5. Factorizing:

   • Since $(u_1 v_2 - u_2 v_1)^2 \geq 0$, we conclude that $\|u\|^2 \|v\|^2 - |u \cdot v|^2 \geq 0$.
   • Therefore, $|u \cdot v|^2 \leq \|u\|^2 \|v\|^2$, completing the proof.

Would you like more details or have any questions about this solution?

Here are five related questions to deepen your understanding:

1. How would this inequality look for vectors in $\mathbb{R}^3$?
2. Can you apply the Cauchy-Schwarz inequality to complex vectors?
3. What are some real-world applications of the Cauchy-Schwarz inequality?
4. How is the dot product related to the angle between two vectors?
5. How would you prove the Cauchy-Schwarz inequality using a geometric approach?

Tip: Remember, the Cauchy-Schwarz inequality is fundamental in many areas of mathematics, including linear algebra, statistics, and physics, as it provides bounds on the relationship between vector magnitudes and their dot products.