The answer is 40. We take an adjustment. Consider the maximum point with most 1 or -1s. By nature of the quadratic functions we find that if x_{i-1}>x_{i+1} then x{i}=(x_{i-1}+x_{i+1})/2, elsewhere x_{i}=1 or -1. It follows that the (x_i)s are composed of some arithmetic progressions from 1 to -1 one after another, for we just consider what is the next term of 1s and what is the next term after -1s. By some final calculations it turns out that the maximum is obtained when there are 3 terms in each arithmetic progression, thus get 40.

Any other solution? Edit: I think I got something. We have $(x_i-1)(x_i+1)(x_{i+1}+1)<=0$, so $x_i^2x_{i+1}<=-x_i^2+x_{i+1}+1$ and analogously, $x_{i+1}^2x_i=>x_{i+1}^2+x_i-1$. So summing gives that the expression is maximum $2(60-x_1^2-x_2^2-...-x_{60}^2)$, if I didn't mess up something. Though, this is probably trivial progress since this is from China TST...

Note the sum is $\frac 13 \sum\limits_{j=1}^{60} (x_j-x_{j+1})^3$ Let $d_j=x_j-x_{j+1}$. We have $\sum d_j=0$ and $d_j\le 2$. We make two observations: If $d_i+d_j\ge 0$ then we should make one $d_i=2$. If $d_i+d_j\le 0$ then we should make $d_i=d_j$. By repeatedly doing the above steps, there is at most one $0<d_j<2$ because if $d_i,d_j$ are both positive I should make one of them 2. This number and the sum of all negative $d_j$'s must be negative, so we make them equal. So this means we have $60-x$ 2's and $x$ numbers equal to $\frac{-2(60-x)}{x}$, we need to maximize $8(60-x)+x(\frac{-2(60-x)}{x})^3 = 8(60-x) - 8\frac{(60-x)^3}{x^2}$ Expanding, we get $480-8x-8x^{-2}(-x^3+180x^2-60^2 \cdot 3x+60^3)$ Means we need to maximize $-8x^{-2} (-60^2 \cdot 3x+60^3)$ or minimize $60^3t^2 - 60^2 \cdot 3t$ where $t=\frac 1x$. We can see that $t=\frac{60^2\cdot 3}{60^3\cdot 2}=\frac{1}{40}$, so $x=40$ and the original sum becomes $40$ as well.

Easy P4.

Cool Problem, similar to others nevertheless posting for storage. Notice that the sum is simply $\frac{1}{3} \sum_{cyc} (x_i-x_{i+1})^3$ Notice for $x \in [-2,2]$, $x^3 \leq 3x+2$ $\frac{1}{3} \sum_{cyc} (x_i-x_{i+1})^3 \leq \frac{1}{3} \sum_{cyc} (3(x_i-x_{i+1})+2) = 60$ achieved by $(1,0,-1,1,0,-1 \cdots ,1,0,-1)$

mathematics2003 wrote: Suppose $x_1,x_2,...,x_{60}\in [-1,1]$ , find the maximum of $$ \sum_{i=1}^{60}x_i^2(x_{i+1}-x_{i-1}),$$where $x_{i+60}=x_i$. If $x_{i+1}-x_{i-1}>=0$ then the equation is convex for $x_i$ so $x_i=-1,1$ (1) For this reson we get that if $x_i=-1$ then $x_{i+1}=-1,1$ and if $x_i=1$ and $x_{i-1}=-1,1$ Now we can replace the sequance $-1,-1,...,-1,1$ with $-1,1$ to win some position similar the $-1,1,1,...,1$ with $-1,1$ So now the sequence became in the form $-1,1,..,-1,1,...,-1,1$ Now in the free spaces we must have $x_{i+2}<x_i$ from (1) Soppuse that the free space has capacity $1$ esily we get that there must be the number $0$ Soppuse that the free space has capacity $2$ esily we get that the best is to be $1/3,-1/3$ which gives $10/27$ Soppuse that the free space has capacity $3$ with the numbers $a,b,c$ we want to maximaize the function: $a-c-a^2-c^3+a^2b-bc^2<a-c-a^2-c^3+a^2+c^2=a-c<2$ Soppuse that the free space has capacity atlest $4$ and let the numbers be $a_1,...,a_l$ then we have: $a_{2}^2(a_3-a_1)+...+a_{l-1}^2(a_{l}-a_{l-2})+a_1-a_l-a_1^2+a_1^2a_2-a_l^2-a_{l-1}a_l^2<a_1-a_l-a_1^2+a_1^2a_2-a_l^2-a_{l-1}a_l^2<=a_1-a_l<=2$ But we can change it with $0,-1,1,0$ and we get value $2$. So the free space has capacity $1$ or $2$ or $3$ suppose that we have $x,y,z$ respectivly then we have that: $x+2y+3z+2(x+y+z)=3x+4y+5z=60$ $A<=2(x+y+z)+y10/27+2z$ Make the chenge $y\to y-3,x\to x+4$ we maximaiz $A$ so $y=0,1,2$ Make also the move $z\to z-2,x\to x+3$ and one more $-1,-1,1$ so $z=0,1$ Making some more moves deling with $y=0,1,2$ and $z=0,1$ we can see that the maximum is when we have $-1,1,0,-1,1,0,...,-1,1,0$. Which gives $40$