TITLE: Help me prove this inequality with positive real numbers
QUESTION [6 upvotes]: Prove that $$p(qr+s+1)(rs+p+1)(sp+q+1)+q(pq+r+1)(rs+p+1)(sp+q+1)+r(pq+r+1)(qr+s+1)(sp+q+1)+s(pq+r+1)(qr+s+1)(rs+p+1)>(pq+r+1)(qr+s+1)(rs+p+1)(sp+q+1)$$ provided that $p,q,r,s$ are positive real numbers for which $p,r>1$ and $q,s<1$. $$ $$ So far I realised that I can divide both sides of the inequality by $(pq+r+1)(qr+s+1)(rs+p+1)(sp+q+1)$ and I got $$\frac{p}{pq+r+1}+\frac{q}{qr+s+1}+\frac{r}{rs+p+1}+\frac{s}{sp+q+1}>1$$ $$$$ I'm not sure how to proceed now or what may lead me to the solution. I will be very grateful for any help on how to solve it.

REPLY [3 votes]: $$
\sum_{cyc} \frac{p}{pq+r+1} = 1+ \sum_{cyc}\left( \frac{p}{pq+r+1} - \frac{p}{p+q+r+s} \right)\\
= 1+ \sum_{cyc}\frac{p(s+p+q-pq-1)}{(pq+r+1)(p+q+r+s)} \\ 
= 1+ \sum_{cyc}\frac{p(s+(p-1)(1-q))}{(pq+r+1)(p+q+r+s)} > 1.\blacksquare
$$