Corollary 2. Suppose that $x_{1}, x_{2}, \ldots, x_{n}$ are non-negative real numbers satisfying
$$x_{1}+x_{2}+\ldots+x_{n}=\text{const}, x_{1}^{2}+x_{2}^{2}+\ldots+x_{n}^{2}=\text{const}$$
and $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ a continuous, symmetric, under-limitary function satisfying that, if $x_{1} \geq x_{2} \geq \ldots \geq x_{n}$ and $x_{2}, x_{3}, \ldots, x_{n-2}$ are fixed then $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)=g\left(x_{1}, x_{n-1}, x_{n}\right)$ is a strictly increasing function of $x_{1} x_{n-1} x_{n}$; then $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ attains the minimum value if and only if $x_{1}=x_{2}=\ldots=x_{k}=0<x_{k+1} \leq x_{k+2}=\ldots=x_{n}$, where $k$ is a certain natural number and $k<n$. If $x_{1} \geq x_{2} \geq \ldots \geq x_{n}$ and $x_{3}, \ldots, x_{n-1}$ are fixed then $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)=g\left(x_{1}, x_{2}, x_{n}\right)$ is a strictly increasing function of $x_{1} x_{2} x_{n}$; then $f\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ attains the maximum value if and only if $x_{1}=x_{2}=\ldots=x_{n-1} \leq x_{n}$.

Proof. To prove the above corollaries, we only show the hardest, that is the second part of the second corollary (and other parts are proved similarly).