Baltic Way 2023 Shortlist

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 $$\prod _{i=1}^n{y}_i^2\ge 1.$$ Thus from the AGM inequality we obtain $$\sum _{i=1}^n\sqrt{y_i}\ge n\prod _{i=1}^n{y}_i^{\frac{1}{2n}}\ge n$$ 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 $$\sum _{i=1}^n\sqrt{(x_i-x_{i-1}{)}^2+y_i}\ge \sqrt{n^2+Y^2}\ge \sqrt{2n^2}=n\sqrt{2}$$ 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

$$\begin{array}{rl}\sum _{i=1}^n\frac{\sqrt{y_i}}{Y}f\left(\frac{x_i-x_{i-1}}{\sqrt{y_i}}\right)& \ge f\left(\sum _{i=1}^n\frac{x_i-x_{i-1}}{Y}\right)=f\left(\frac{n}{Y}\right),\text{ i.e.}\\ \sum _{i=1}^n\sqrt{(x_i-x_{i-1}{)}^2+y_i}& \ge \sqrt{n^2+Y^2}\end{array}$$ 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 $$withequalityiff$y_i = y_{i+1}$and$y_i y_{i+1} = 1$forall$i$,i.e.$y_i = 1$forall$i$.