Animated Solution for Mathematics - Sequence and Series: If $x_1,x_2,\dots ,x_n$ are any real numbers and $n$ is any positive integer, then

Welcome, future engineer! Today, we are going to peel back the curtain on one of the most elegant and powerful tools in the mathematician's toolkit: the Cauchy-Schwarz Inequality.

When you first look at a problem asking you to relate ${\sum }_{i=1}^n{x}_i^2$ and $({\sum }_{i=1}^n{x}_i{)}^2$, it is easy to feel overwhelmed. It looks like a mess of summations and squares, but this is a structure waiting to be revealed.

Before we touch a single variable, let's visualize what is happening. Imagine two vectors, $\vec{a}$ and $\vec{b}$, floating in $n$-dimensional space.

The Cauchy-Schwarz inequality tells us that the square of their dot product, $(\vec{a}\cdot \vec{b}{)}^2$, can never exceed the product of their squared magnitudes, $|\vec{a}{|}^2|\vec{b}{|}^2$. This is a profound geometric truth stating that the 'alignment' of two vectors is limited by their individual lengths.

In algebraic terms, this translates to the beautiful inequality:

This is our weapon of choice.

Now, here is where the magic happens. We have the inequality, but we need to make it fit our specific problem.

We are looking at the term $({\sum }_{i=1}^n{x}_i{)}^2$. If we look at the left side of our Cauchy-Schwarz formula, $({\sum }_{i=1}^n{a}_i{b}_i{)}^2$, we see that if we set $a_i=x_i$, we are halfway there.

To keep $x_i$ unchanged while multiplying it by something, we must multiply it by $1$. So, we make the strategic choice: let $a_i=x_i$ and $b_i=1$ for every index $i$ from $1$ to $n$.

Let's watch the pieces fall into place. On the left side, we have $({\sum }_{i=1}^n{x}_i\cdot 1{)}^2$, which simplifies perfectly to $({\sum }_{i=1}^n{x}_i{)}^2$.

On the right side, we have the product of two sums. The first is ${\sum }_{i=1}^n{a}_i^2$, which becomes ${\sum }_{i=1}^n{x}_i^2$. The second is ${\sum }_{i=1}^n{b}_i^2$, which becomes ${\sum }_{i=1}^n{1}^2$.

Since $1^2$ is just $1$, we are summing $1$ exactly $n$ times, which gives us $n$. Thus, the right side becomes $n\cdot {\sum }_{i=1}^n{x}_i^2$. Putting it all together, we get the fundamental inequality:

Now, look at the options provided in the problem. We derived the relationship:

Compare this to the choices. If the options provided do not match this derivation, trust your work. You have successfully navigated the logic, applied a high-level theorem, and arrived at the mathematical truth.