, 2023

We claim that the minimal value of the desired expression is $n\sqrt{2}$ achieved by $x_i = i$ and $y_i = 1$ for all $i$. In what follows below, $Y$ denotes the sum $Y = \sum_{i=1}^{n} \sqrt{y_i}$.
The solution consists of proving two separate inequalities.

By taking the global product of all the inequalities of the form $y_i y_{i+1} \ge 1$ and $y_n y_1 \ge 1$ we obtain
$$\prod_{i=1}^{n} y_i^2 \geq 1.$$
Thus from the AGM inequality we obtain
$$\sum_{i=1}^{n} \sqrt{y_i} \geq n \prod_{i=1}^{n} y_i^{\frac{1}{2n}} \geq n$$
with equality in the left-hand inequality for equal $y_i$ and in the right-hand inequality for $y_i = 1$.
Hence, by combining the two inequalities we obtain
$$\sum_{i=1}^{n} \sqrt{(x_i - x_{i-1})^2 + y_i} \geq \sqrt{n^2 + Y^2} \geq \sqrt{2n^2} = n\sqrt{2}$$
with equality iff $y_i = 1$ for all $i$ and $x_i = x_{i-1} + 1$, i.e. $x_i = i$.

2nd Solution:
For the two inequalities we give a second proof, each.
First Inequality:
Again we consider the $y_i$ as constants, and rewrite
$$\sqrt{(x_i - x_{i-1})^2 + y_i} = \sqrt{y_i} \sqrt{1 + \left(\frac{x_i - x_{i-1}}{\sqrt{y_i}}\right)^2}.$$
By considering the second derivative $\frac{\partial^2}{\partial x^2} \sqrt{1+x^2} = \frac{1}{(1+x^2)^{\frac{3}{2}}} > 0$ for all $x$ we find that $f(x) = \sqrt{1+x^2}$ is a convex function.
Hence, by applying the weighted Jensen's Inequality with weights $\frac{\sqrt{y_i}}{Y}$ we obtain

$$\begin{aligned} \sum_{i=1}^{n} \frac{\sqrt{y_i}}{Y} f\left(\frac{x_i - x_{i-1}}{\sqrt{y_i}}\right) &\ge f\left(\sum_{i=1}^{n} \frac{x_i - x_{i-1}}{Y}\right) = f\left(\frac{n}{Y}\right), \text{ i.e.} \\ \sum_{i=1}^{n} \sqrt{(x_i - x_{i-1})^2 + y_i} &\ge \sqrt{n^2 + Y^2} \end{aligned}$$
$$with equality if and only if \$\frac{x_i - x_{i-1}}{\sqrt{y_i}}\$ is independent of \$i\$. Hence the equality case may be established identically to the first solution. **Second Inequality:** By rewriting the given inequalities as \$\sqrt[4]{y_i y_{i+1}} \ge 1\$, taking the global sum and using the Rearrangement Inequality we obtain$$
$\sum_{i=1}^{n} \sqrt{y_i} \ge \sum_{i=1}^{n} \sqrt[4]{y_i y_{i+1}} \ge$ n
$$$$
with equality iff $y_i = y_{i+1}$ and $y_i y_{i+1} = 1$ for all $i$, i.e. $y_i = 1$ for all $i$.