# Proving that $f(x_1,x_2\ldots x_n)=\frac{1}{n}\sum_{i}x_i^2 - \frac{1}{n^2}\sum_{i\neq j}\ln|x_i-x_j|\geq 4$

I came across this inequality in Andersen, Giounnet and Zeitouni's book on Random Matrix Theory. Now, we are asked to establish this using the two inequalities $\ln|x_i-x_j|\leq \ln(|x_i|+1)+\ln(|x_j|+1)$ and $x^2-2\ln(1+|x|)\geq -4$. Further, we are required to verify that $-\frac{1}{n^2}\sum_{i\neq j}\ln|x_i-x_j|\leq f(x_1,x_2\ldots x_n)+4$. What I am able to verify directly is that $f(x_1,x_2\ldots x_n)\geq -4$, but not sure how to proceed. For context, this is involved in the proof that Dyson Brownian Motion never hits the boundary of the Weyl Chamber $x_1<x_2<x_3\ldots <x_n$. Any resources towards that end will also be helpful.

• The hint of the book kind of makes me think that they want us to use induction because they have given us the base case. Have you tried induction? Moreover, is $N=n$? If not, what is $N$? – stressed out Dec 7 '17 at 10:08
• I couldn't really see why induction was necessary for the first inequality anyway. If the logarithm $\ln|x_i-x_j|$ is split as above, we obtain a sum where the $i^{th}$ term depends only on $x_i$, which we can bound below as above. Do you mean induction for the second inequality? – Kesav Krishnan Dec 7 '17 at 10:12
• No, I meant induction for your final inequality about $f(x_1,x_2,\cdots,x_n) \geq 4$ assuming those two inequalities. What is $N$? – stressed out Dec 7 '17 at 10:17
• Please fix it in the question as well. Also, $f(x_1,\cdots,x_n)\geq 4$ or $-4$? – stressed out Dec 7 '17 at 10:23
• What I've been able to find is that $f(x_1\ldots x_n) \geq -4$. The proof seems to rely on $f\geq 4$.The inequality $-\frac{1}{n^2}\sum_{i\neq j}\ln|x_i-x_j|\leq f(x_1\ldots x_n) +4$ is also used in the proof, but I am not able to verify this myself – Kesav Krishnan Dec 7 '17 at 10:29