no proof

JoeBlow wrote:

no proof

After solving it I understand what you mean! I solved it making extensive use of mixing variables, to reduce it to maximizing a function defined for each partition of n.

We have a continuos function on a compact set $ [0,1]^n$, hence there is an optimal point $ (x_1,...,x_n)$. Note now that 0) impossible to have $ x_{i - 1} = x_{i} = x_{i + 1}$; 1) if $ x_i\leq x_{i - 1}$ and $ x_i\leq x_{i + 1}$, then $ x_i = 0$; 2) if $ x_i\geq x_{i - 1}$ and $ x_i\geq x_{i + 1}$, then $ x_i = 1$; 3) if $ x_{i + 1}\leq x_i\leq x_{i - 1}$ or $ x_{i - 1}\leq x_i\leq x_{i + 1}$, then $ x_i = \frac {x_{i - 1} + x_{i + 1}}{2}$. It follows that $ (x_1,...,x_n)$ looks like \[ (0,\frac {1}{k_1},\frac2{k_1},...,1,\frac {k_2 - 1}{k_2},...,\frac2{k_2},\frac1{k_2},0,\frac1{k_3},...,\frac1{k_l}), \] where $ k_1$, $ k_2$, ..., $ k_l$ are natural numbers, $ k_1 + k_2 + ... + k_l = n$, $ l$ is even clearly. Then the function is this point equals \[ S = \sqrt {k_1} + \sqrt {k_2} + ...\sqrt {k_l}. \] Using the fact that $ l$ is even and $ \sqrt {k} < \sqrt {k - 1} + 1$ we conclude that maximal possible value of $ S$ is $ n - 2 + \sqrt 2$ ($ l = n - 1$, $ k_1 = k_2 = ... = k_{l - 1} = 1$, $ k_l = 2$ in this case). It is not a real olympiad problem in sense of inner beauty.

since $ n$ is odd, there must be an $ i$ such that both $ x_i$ and $ x_{i + 1}$ are both belong to $ [0,\frac {1}{2}]$ or $ [\frac {1}{2},1]$. without loss of generality let $ x_1\leq x_2$ and $ x_1$, $ x_2$ belong to $ [0,\frac {1}{2}]$. We can prove that $ \sqrt {x_2 - x_1} + \sqrt {Ix_3 - x_2I}\leq \sqrt {2}$. If $ x_3 > x_2$, $ \sqrt {x_2 - x_1} + \sqrt {x_3 - x_2}\leq 2\cdot \sqrt {\frac {x_3 - x_1}{2}}\leq \sqrt {2}$; else $ x_1,x_2,x_3$ are all belong to $ [0,\frac {1}{2}]$. Hence, $ \sqrt {x_2 - x_1} + \sqrt {Ix_3 - x_2I}\leq \sqrt \frac {1}{2} + \sqrt \frac {1}{2}$. Also all of the other terms of the sum are less then or equal to $ 1$. summing them gives the desired result. example is $ (0,\frac {1}{2},1,0,1,\ldots ,1)$ note: all the indices are considered in modulo $ n$

We can assume there exists some $ i$ for which $ x_i \leq x_{i+1} \leq x_{i+2}$ because if it didnt exist: we can WLOG suppose $ x_1=max{x_i}$. Now $ x_1 \geq x_2$. It follows that $ x_2 \leq x_3$ because if not our assumption would be faulty. Now we continue, it would be: $ x_1 \geq x_2 \leq x_3 \geq x_4 .... \leq x_n \geq x_1$, because n is odd. It's a contradiction that $ x_1=max{x_i}$. So there exists $ i$ such that $ x_i \leq x_{i+1} \leq x_{i+2}$. Now we can use AM-QM: $ \sqrt{_x{i+2}-x_{i+1}}+\sqrt{x_{i+1}-x_i} \leq \sqrt{2}\sqrt{x_{i+2}-x_i}$. Now, maximum is attained if all roots are equal to $ 1$. Now it easy to check this is attainable, if $ i$ is even, then $ (x_1,x_2,...,x_i,x_{i+1},x_{i+2},...,x_n)=(0,1,...,1,\frac{1}{2},0,...,1)$. Similarly if $ i$ is odd. The maximum is $ n-2+\sqrt{2}$.

I think that this is pretty similar to IMO 2021/2. Here we can use shifting to let the biggest number be $1$ or the smallest to be $0$, and then induction will probably finish.