Amir Hossein wrote: Prove for all positive real numbers $m,n,p,q$ that $$\frac{m}{t+n+p+q} + \frac{n}{t+p+q+m} + \frac{p}{t+q+m+n} + \frac{q}{t+m+n+p} \geq \frac{4}{5},$$where $t=\frac{m+n+p+q}{2}.$ Write your inequality become \[\sum_{m,n,p,q}\frac{m}{m+3n+3p+3q} \geqslant \frac25.\]From Cauchy-Schwarz inequality, we have \[\sum_{m \to n \to p \to q}\frac{m}{m+3n+3p+3q} \geqslant \frac{(m+n+p+q)^2}{\displaystyle \sum_{m \to n \to p \to q} m(m+3n+3p+3q)}.\]So we will show that \[5(m+n+p+q)^2 \geqslant 2\sum_{m \to n \to p \to q} m(m+3n+3p+3q).\]or \[3m^2-2mn-2mp-2mq+3n^2-2np-2nq+3p^2-2pq+3q^2 \geqslant 0.\]Which is Maclaurin inequality. So we are done.

This is posted in the wrong contest collection (2018 and it is from 2017)