I don't understand how you expanded the quadratic expression?

Certainly! Let's carefully go through the expansion of the quadratic expression.

### Given Expression:
We start with the expression given in the problem:
\[
(t\vec{x} + \vec{y})(t\vec{x} + \vec{y})^\top \geq 0.
\]
Here, \( t \) is a scalar, and \(\vec{x}\) and \(\vec{y}\) are row vectors in \(\mathbb{R}^n\). The notation \((t\vec{x} + \vec{y})^\top\) represents the transpose of the row vector \( (t\vec{x} + \vec{y}) \), making it a column vector.

### Expansion of the Expression:
Let's expand \((t\vec{x} + \vec{y})(t\vec{x} + \vec{y})^\top\) step by step.

1. **First Term**: 
   \[
   t\vec{x}(t\vec{x})^\top = t\vec{x} \cdot t\vec{x}^\top = t^2 (\vec{x}\vec{x}^\top).
   \]
   Here, \(\vec{x}\vec{x}^\top\) is a scalar (since it's a row vector multiplied by its transpose).

2. **Second Term**:
   \[
   t\vec{x} \cdot \vec{y}^\top.
   \]
   This is a scalar since it is the dot product of two vectors.

3. **Third Term**:
   \[
   \vec{y} \cdot (t\vec{x})^\top = t(\vec{y} \cdot \vec{x}^\top).
   \]
   Again, this is a scalar.

4. **Fourth Term**:
   \[
   \vec{y}\vec{y}^\top.
   \]
   This term is also a scalar.

### Full Expansion:
Now, let's put it all together:

\[
(t\vec{x} + \vec{y})(t\vec{x} + \vec{y})^\top = t^2 (\vec{x}\vec{x}^\top) + t (\vec{x}\vec{y}^\top) + t (\vec{y}\vec{x}^\top) + \vec{y}\vec{y}^\top.
\]

But note that \( \vec{x}\vec{y}^\top \) and \( \vec{y}\vec{x}^\top \) are actually the same scalar, so we can combine these terms:

\[
= t^2 (\vec{x}\vec{x}^\top) + 2t (\vec{x}\vec{y}^\top) + \vec{y}\vec{y}^\top.
\]

This is a quadratic expression in \( t \), with:
- \( a = \vec{x}\vec{x}^\top \) (coefficient of \( t^2 \)),
- \( b = 2\vec{x}\vec{y}^\top \) (coefficient of \( t \)),
- \( c = \vec{y}\vec{y}^\top \) (constant term).

### Summary:
Thus, the expression \((t\vec{x} + \vec{y})(t\vec{x} + \vec{y})^\top\) expands into the quadratic form:
\[
t^2 (\vec{x}\vec{x}^\top) + 2t (\vec{x}\vec{y}^\top) + \vec{y}\vec{y}^\top \geq 0.
\]
This is the basis for using the discriminant to establish the inequality.

Would you like to explore further or clarify any other steps?

### Related Questions:
1. Why does the quadratic expression need to be non-negative?
2. How do the vectors \(\vec{x}\) and \(\vec{y}\) interact geometrically in this inequality?
3. What role does the discriminant play in determining the nature of quadratic equations?
4. Can this proof technique be applied to other forms of inequalities?
5. What is the significance of the Cauchy-Schwarz inequality in inner product spaces?

### Tip:
When expanding expressions like this, carefully track each term and how they combine. Visualizing vector multiplication can help you understand the nature of each term.

Learn how to expand the quadratic expression (t\vec{x} + \vec{y})(t\vec{x} + \vec{y})^\top with a detailed step-by-step solution. This problem involves vector and matrix algebra, including the application of the Cauchy-Schwarz inequality. Suitable for advanced high school students and those interested in advanced algebraic techniques.

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