Given $n \geq 2$ reals $x_1 , x_2 , \dots , x_n.$ Show that $$\prod_{1\leq i < j \leq n} (x_i - x_j)^2 \leq \prod_{i=0}^{n-1} \left(\sum_{j=1}^{n} x_j^{2i}\right)$$and find all the $(x_1 , x_2 , \dots , x_n)$ where the equality holds.
Problem
Source: IMOC 2021 A11
Tags: inequalities
BlueCloud12
11.08.2021 16:40
when you take half of them a and half of them -a this is not necessarily right for all a>0. However for positive reals thie inequality certainly holds.
pi_quadrat_sechstel
11.08.2021 20:41
BlueCloud12 wrote: when you take half of them a and half of them -a this is not necessarily right for all a>0. However for positive reals thie inequality certainly holds. This is not a counterexample. For $n=2$ we have equality and for $n\geq3$ the LHS is 0 and the RHS nonnegative.
Rickyminer
14.08.2021 17:20
It is just Hadamard inequality for Vandermonde matrix \[ \begin{pmatrix} 1 & x_1 & \cdots & x_1^{n-1} \\ 1 & x_2 & \cdots & x_2^{n-1} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_n & \cdots & x_n^{n-1} \end{pmatrix}. \]The equality holds only when its column vectors are pairwise orthogonal. It implies that $x_1^k+x_2^k+\cdots+x_n^k=0$ for $k=1,2,\ldots,2n-3$. If $n=2$, we have $x_1+x_2=0$. If $n \ge 3$, we prove that $x_1=x_2=\cdots=x_n=0$.
We only need $x_1^k+x_2^k+\cdots+x_n^k=0$ for $k=1,2,\ldots,n$ (note that $2n-3 \ge n$). Observe that we can simply omit $x_i=0$, as it reduces to case $n-1$. Assume that there is a non-zero solution, let the number of pairwise different $x_i$ be $r$. Uniting the equal $x_i$, we get
\[ m_1 y_1^k+\cdots+m_r y_r^k=0 ~,~ \text{for} ~ k=1,2,\ldots,n. \]Set $z_j=m_j y_j, j=1,2,\ldots,r$, the equations can be written as
\[ z_1 y_1^{k-1}+\cdots+z_r y_r^{k-1}=0 ~,~ \text{for} ~ k=1,2,\ldots,n. \]Pick the first $r$ equations, we get a system of linear equations for $z_1,\ldots,z_r$, whose determinant is the Vandermonde determinant $V(y_1,\ldots,y_r) \neq 0$. Hence it only has trivial solution $z_1=\cdots=z_r=0$, therefore $y_1=\cdots=y_r=0$. The contradiction implies that there is only the zero solution.
Indeed, $x_1^2+x_2^2+\cdots+x_n^2=0$ implies $x_1=x_2=\cdots=x_n=0$.
Untro368
15.08.2021 09:46
Rickyminer wrote: It is just Hadamard inequality for Vandermonde matrix \[ \begin{pmatrix} 1 & x_1 & \cdots & x_1^{n-1} \\ 1 & x_2 & \cdots & x_2^{n-1} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & x_n & \cdots & x_n^{n-1} \end{pmatrix}. \]The equality holds only when its column vectors are pairwise orthogonal. It implies that $x_1^k+x_2^k+\cdots+x_n^k=0$ for $k=1,2,\ldots,2n-3$. If $n=2$, we have $x_1+x_2=0$. If $n \ge 3$, we prove that $x_1=x_2=\cdots=x_n=0$.
We only need $x_1^k+x_2^k+\cdots+x_n^k=0$ for $k=1,2,\ldots,n$ (note that $2n-3 \ge n$). Observe that we can simply omit $x_i=0$, as it reduces to case $n-1$. Assume that there is a non-zero solution, let the number of pairwise different $x_i$ be $r$. Uniting the equal $x_i$, we get
\[ m_1 y_1^k+\cdots+m_r y_r^k=0 ~,~ \text{for} ~ k=1,2,\ldots,n. \]Set $z_j=m_j y_j, j=1,2,\ldots,r$, the equations can be written as
\[ z_1 y_1^{k-1}+\cdots+z_r y_r^{k-1}=0 ~,~ \text{for} ~ k=1,2,\ldots,n. \]Pick the first $r$ equations, we get a system of linear equations for $z_1,\ldots,z_r$, whose determinant is the Vandermonde determinant $V(y_1,\ldots,y_r) \neq 0$. Hence it only has trivial solution $z_1=\cdots=z_r=0$, therefore $y_1=\cdots=y_r=0$. The contradiction implies that there is only the zero solution.
Indeed, $x_1^2+x_2^2+\cdots+x_n^2=0$ implies $x_1=x_2=\cdots=x_n=0$. Actually, for $n>2$ you can use Newton identity to get \[\sum_{\text{sym}}{x_1\cdots x_j}=0\ \forall j=1,...,n\]and thus the polynomial with $x_1,...,x_n$ be roots is $x^n=0$ i.e. $x_i=0$ for each $i$.