$$\left(\frac{1+y\left(a^pb^qc^r\right)^{\frac{1}{p+q+r}}}{1+x\left(a^pb^qc^r\right)^{\frac{1}{p+q+r}}}\right)^{\frac{p+q+r}{\left(a^pb^qc^r\right)^{\frac{1}{p+q+r}}}}\left(\frac{1+xa}{1+ya}\right)^{\frac{p}{a}}\left(\frac{1+xb}{1+yb}\right)^{\frac{q}{b}}\left(\frac{1+xc}{1+yc}\right)^{\frac{r}{c}}\geq \prod \limits_{cyc}\left(\frac{1+y\left(a^pb^q\right)^{\frac{1}{p+q}}}{1+x\left(a^pb^q\right)^{\frac{1}{p+q}}}\right)^{\frac{p+q}{\left(a^pb^q\right)^{\frac{1}{p+q}}}}$$
This is a inequality problem from 2019 Jozsef Wildt International Mathematics Competition. For this problem i almost tried in all ways with AM-GM and Cauchy-Schwarz but i am unable to do it. Honestly i don't even know how to start thinking for this problem. Please help. I know i have not uploaded my attempts. But what i shall attempt. I am completely blank about this problem
We need to prove that $g(y)\geq0,$ where $$g(y)=\tfrac{p+q+r}{\left(a^pb^qc^r\right)^{\tfrac{1}{p+q+r}}}\ln\tfrac{1+y\left(a^pb^qc^r\right)^{\tfrac{1}{p+q+r}}}{1+x\left(a^pb^qc^r\right)^{\tfrac{1}{p+q+r}}}+\sum_{cyc}\tfrac{p}{a}\ln\tfrac{1+ya}{1+xa}-\sum_{cyc}\tfrac{p+q}{\left(a^pb^q\right)^{\tfrac{1}{p+q}}}\ln\tfrac{1+y\left(a^pb^q\right)^{\tfrac{1}{p+q}}}{1+x\left(a^pb^q\right)^{\tfrac{1}{p+q}}}.$$ But $$g'(y)=\tfrac{p+q+r}{1+y\left(a^pb^qc^r\right)^{\tfrac{1}{p+q+r}}}+\sum_{cyc}\tfrac{p}{1+ya}-\sum_{cyc}\tfrac{p+q}{1+y\left(a^pb^q\right)^{\tfrac{1}{p+q}}}=$$ $$=\frac{p+q+r}{1+ye^{\frac{\sum\limits_{cyc}p\ln{a}}{p+q+r}}}+\sum_{cyc}\frac{p}{1+ye^{\ln{a}}}-\sum_{cyc}\frac{p+q}{1+ye^{\frac{p\ln{a}+q\ln{b}}{p+q}}}.$$ Now, let $f(t)=\frac{1}{1+ye^t},$ where $t\geq0$ and $y\geq1$.
Thus, $$f''(t)=-\left(\frac{ye^t}{(1+ye^t)^2}\right)'=-\left(\frac{ye^t+1-1}{(1+ye^t)^2}\right)'=$$ $$=\frac{ye^t}{(1+ye^t)^2}-\frac{2ye^t}{(1+ye^t)^3}=\frac{ye^t(ye^t-1)}{(1+ye^t)^3}\geq0,$$ which says that $f$ is a convex function.
Thus, by the Tiberiu Popoviciu's inequality $g'(y)\geq0,$ which says $$g(y)\geq g(x)=0$$ and we are done!
About Popoviciu see here: https://arxiv.org/pdf/0803.2958v1.pdf