We claim the minimum value is $1+2+...+n$ or $\frac{n(n+1)}{2}$ obtained when $x_i = i$ Firstly,... $\bold {Claim \:1:}$ $x_1^3+\dots+x_n^3 \geq (x_1+\dots+x_n)^2$ We will prove this inductively, $n = 1, 2$ is trivial W.L.O.G $1\leq x_1 < x_2 < ... < x_n$ Assuming $x_1^3+\dots+x_n^3 \geq (x_1+\dots+x_n)^2$ is true, we will prove $x_1^3+\dots+x_{n+1}^3 \geq (x_1+\dots+x_{n+1})^2$ We will have $x_1^3+\dots+x_{n+1}^3 \geq (x_1+\dots+x_n)^2 + x_{n+1}^3$ This implies $x_1^3+\dots+x_{n+1}^3 \geq (x_1+\dots+x_n)^2 + x_{n+1}^3$ We have to show that $(x_1+\dots+x_n)^2 + x_{n+1}^3 \geq (x_1+\dots+x_{n+1})^2$ Or $$x_{n+1}^2 \geq 2(x_1+\dots+x_n) + x_{n+1}$$ Obviously equality happens when $x_i = i$. Shift the indices value by $k$, for arbitrary positive value of k such that for every i, we have $x_i = i+k$ Then from the inequality, we have $$(n+1+k)^2 \geq 2[(1+k)+(2+k)...+(n+k)] + n+1+k$$$$\implies (n+1+k)^2 \geq 2[\frac{(n+k)(n+k+1)}{2} - \frac{k(k+1)}{2}] + n+1+k$$ Which obviously true for all positive integers $k$, since $$(n+1+k)^2 \geq (n+k)(n+k+1) - k(k+1) + (n+1+k)$$$$\implies (n+1+k)^2 \geq (n+k+1)^2 - k(k+1)$$ This implies $x_{n+1}^2 \geq 2(x_1+\dots+x_n) + x_{n+1}$ true for $1\leq x_1 < x_2 < ... < x_n$ because for every random shifting, other than mentioned above, makes the RHS smaller than what we have proven, which is obviously true. Hence proven. Notice that the claim implies that is the minimum value since we have checked every possible value of $x_1, x_2, ..., x_n$ where $1\leq x_1 < x_2 < ... < x_n$ Then we have $$\frac{x_1^3+\dots+x_n^3}{x_1+\dots+x_n} \geq \frac{(x_1+\dots+x_n)^2}{x_1+\dots+x_n} \geq x_1+\dots+x_n \geq 1+2+...+n = \frac{n(n+1)}{2}$$

is this okay? we claim that the answer is n(n+1)/2 it can be reach by Xi=i i=1~n denote f(X1,X2....)to represent the function we use the method of adjustment and claim that f(1,2,3,4.....n) is the smallest proof: if there is a m>n among X1~Xn then we have: f(1,2,3....i-1,i+1,i+2....n,m)-f(1,2,.....n)= [1^3+2^3+....+n^3+m^3-i^3] / 1+2+....+n+m-i -n(n+1)/2 (??>0) after some easy algebra calculation we ill get 2(m^2+mi+i^2)>n(n+1) and this is trivial this is the edition of changeing 1 element from 1,2,.....n it is easy to see that changing any amount of element in 1,2,.....n still holds the inequality (by using 2(m^2+mi+i^2)>n(n+1) for many times for different m,i) so we finish our proof