Maximal (minimal) value of $S=x_1^2x_2+x_2^2x_3+\cdots+x_{n-1}^2x_n+x_n^2x_1$ on condition that $x_1^2+x_2^2+\cdots+x_n^2=1$ since $x_1^2+x_2^2+\cdots+x_n^2=1$ is sphere,a compact set,so $S$ has a maximal(minimal) value. But when I try to solve it using the Lagrangian multiplier method, I don't know how to solve these equations. Clearly $x_1=x_2=x_3=\cdots = x_n=\frac{1}{\sqrt{n}}$ is an extremal point, but I don't know if it's the maximal(minimal) value.
I want to know how to solve the problem. Also, could this problem be solved by elementary methods, like some inequality  techniques?
 A: For $n=3$ we need to find $$\max_{a^2+b^2+c^2=1}(a^2b+b^2c+c^2a).$$
Indeed, let $\{|a|,|b|,|c|\}=\{x,y,z\}$, where $x\geq y\geq z\geq0$.
Thus, by Rearrangement and AM-GM we obtain:
$$\sum_{cyc}a^2b\leq|a|\cdot(|a||b|)+|b|\cdot(|b||c|)+|c|\cdot(|c||a|)\leq$$
$$\leq x\cdot xy+y\cdot xz+z\cdot yz=y(x^2+xz+z^2)\leq y\left(x^2+\frac{x^2+z^2}{2}+z^2\right)=$$
$$=\frac{3}{2}y(1-y^2)=\frac{3}{2\sqrt2}\sqrt{2y^2(1-y^2)^2}\leq\frac{3}{2\sqrt2}\sqrt{\left(\frac{2y^2+2-2y^2}{3}\right)^3}=\frac{1}{\sqrt3}.$$
The equality occurs for $a=b=c=\frac{1}{\sqrt3}$, which says that we got a maximal value.
For $n=4$ we need to find $$\max_{\sum\limits_{cyc}a^2=1}\sum_{cyc}a^2b.$$
Indeed, by C-S and AM-GM we obtain:
$$\sum_{cyc}a^2b\leq\sqrt{\sum_{cyc}a^2\sum_{cyc}a^2b^2}=\sqrt{(a^2+c^2)(b^2+d^2)}\leq\frac{1}{2}(a^2+c^2+b^2+d^2)=\frac{1}{2}.$$
The equality occurs for $a=b=c=d=\frac{1}{2},$ which says that we got a maximal value.
For $n\geq5$ we can use the Lagrange Multipliers method, but it does not give nice numbers.
For example, for $n=5$ the maximum occurs, when $(x_1,x_2,x_3,x_4,x_5)||(0.79...,3.24...,3.78...,2.48...,1),$ which gives a value $0.45...$
The following inequality is also true.

Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=1$. Prove that:
$$a^3b^2+b^3c^2+c^3a^2\leq\frac{1}{3\sqrt3}.$$

