Please make sure the latex for the following text is rendering properly for me also focus on the triangle inequality and give me something I can use for my notes, step by step: Certainly! Here we’ll go in detail through the key components of the mentioned text from a linear algebra standpoint, focusing on the Cauchy-Schwarz Inequality and the Triangle Inequality, which are foundational concepts in studying vectors and their properties.

### Theorem 3.2: Cauchy-Schwarz Inequality

#### Statement
For any vectors ( \mathbf{u} ) and ( \mathbf{v} ) in ( \mathbb{R}^n ),
[ (\mathbf{u} \cdot \mathbf{v})^2 \leq |\mathbf{u}|^2 |\mathbf{v}|^2 ]

#### Explanation and Proof
The dot product of two vectors ( \mathbf{u} ) and ( \mathbf{v} ), denoted as ( \mathbf{u} \cdot \mathbf{v} ), can be expanded (in an orthonormal coordinate system) as:
[ \mathbf{u} \cdot \mathbf{v} = \sum_{i=1}^n u_i v_i ]

where ( u_i ) and ( v_i ) are the components of ( \mathbf{u} ) and ( \mathbf{v} ), respectively.

We begin by considering a real-valued function for a real variable ( t ):
[ f(t) = |\mathbf{v}t - \mathbf{u}|^2 ]
Expanding this, we get:
[ f(t) = (\mathbf{v}t - \mathbf{u}) \cdot (\mathbf{v}t - \mathbf{u}) = |\mathbf{v}|^2 t^2 - 2 (\mathbf{u} \cdot \mathbf{v}) t + |\mathbf{u}|^2 ]
Since ( f(t) \geq 0 ) for all ( t ) (as it represents the squared norm of a vector, which is always non-negative), the discriminant of this quadratic equation must be non-positive for the equation to have no real positive roots (which ensures ( f(t) \geq 0 )):
[ 4 (\mathbf{u} \cdot \mathbind{v})^2 - 4 |\mathbf{u}|^2 |\mathbf{v}|^2 \leq 0 ]
Dividing by 4 and rearranging gives:
[ (\mathbf{u} \cdot \mathbf{v})^2 \leq |\mathbf{u}|^2 |\mathbf{v}|^2 ]

### Theorem 3.3: Triangle Inequality

#### Statement
For any vectors ( \mathbf{u} ) and ( \mathbf{v} ) in ( \mathbb{R}^n ),
[ |\mathbf{u} + \mathbf{v}| \leq |\mathbf{u}| + |\mathbf{v}| ]

#### Explanation and Proof
Square both sides of the inequality:
[ |\mathbf{u} + \mathbf{v}|^2 \leq (|\mathbf{u}| + |\mathbf{v}|)^2 ]
Expanding both sides, we get:
[ |\mathbf{u} + \mathbf{v}|^2 = (\mathbf{u} + \mathbf{v}) \cdot (\mathbf{u} + \mathbf{v}) = |\mathbf{u}|^2 + 2 (\mathbf{u} \cdot \mathbf{v}) + |\mathbf{v}|^2 ]
[ (|\mathbf{u}| + |\mathbf{v}|)^2 = |\mathbf{u}|^2 + 2 |\mathbf{u}| |\mathbf{v}| + |\mathbf{v}|^2 ]
Using the fact from the Cauchy-Schwarz Inequality that ( |(\mathbf{u} \cdot \mathbf{v})| \leq |\mathbf{u}| |\mathbf{v}| ), we know:
[ |\mathbf{u}|^2 + 2 (\mathbf{u} \cdot \mathbf{v}) + |\mathbf{v}|^2 \leq |\mathbf{u}|^2 + 2 |\mathbf{u}| |\mathbf{v}| + |\mathbf{v}|^2 ]
Thus, we confirm that:
[ |\mathbf{u} + \mathbf{v}|^2 \leq (|\mathbf{u}| + |\mathbf{v}|)^2 ]
Taking square roots on both sides (keeping in mind these are norms and are non-negative), we obtain the desired inequality:
[ |\mathbf{u} + \mathbf{v}| \leq |\mathbf{u}| + |\mathbf{v}| ]

Understanding these theorems geometrically also contributes to their intuitive grasp: The Cauchy-Schwarz Inequality corresponds to the notion that the projection of one vector onto another cannot exceed the vectors’ magnitudes, and the Triangle Inequality is akin to the rule in Euclidean geometry that the sum of lengths of two sides of a triangle is always greater than the length of the third side.