Belarusian Mathematical Olympiad

The required inequality $$\frac{xyzt}{(x+y)(z+t)}\le \frac{(x+z{)}^2(y+t{)}^2}{4(x+y+z+t{)}^2}$$ is equivalent to the inequality $$\frac{(x+y)(z+t)}{xyzt}\ge \frac{4(x+y+z+t{)}^2}{(x+z{)}^2(y+t{)}^2}$$ (1) Note that $$\frac{2(x+y+z+t)}{(x+z)(y+t)}=\frac{2}{x+z}+\frac{2}{y+t}$$ By A.M. - G.M. inequality, $$\frac{2}{x+z}\le \frac{1}{\sqrt{xz}},\frac{2}{y+t}\le \frac{1}{\sqrt{yt}}$$ hence $$\frac{4(x+y+z+t{)}^2}{(x+z{)}^2(y+t{)}^2}\le {\left(\frac{1}{\sqrt{xz}}+\frac{1}{\sqrt{yt}}\right)}^2$$ Further, $$\frac{(x+y)(z+t)}{xyzt}=\frac{xz+xt+yz+yt}{xyzt}$$ $$=\frac{xz}{xyzt}+\frac{xt}{xyzt}+\frac{yz}{xyzt}+\frac{yt}{xyzt}$$ $$=\frac{1}{yt}+\frac{1}{yz}+\frac{1}{xt}+\frac{1}{xz}$$ But we can also write $$\frac{(x+y)(z+t)}{xyzt}=\frac{xz+xt+yz+yt}{xyzt}=\frac{xz+yt+xt+yz}{xyzt}$$ $$=\frac{xz+yt}{xyzt}+\frac{xt+yz}{xyzt}$$ $$=\frac{xz}{xyzt}+\frac{yt}{xyzt}+\frac{xt}{xyzt}+\frac{yz}{xyzt}$$ $$=\frac{1}{yt}+\frac{1}{xz}+\frac{1}{xt}+\frac{1}{yz}$$ But the key step is to use the AM-GM inequality: $$\frac{(x+y)(z+t)}{xyzt}\ge {\left(\frac{1}{\sqrt{xz}}+\frac{1}{\sqrt{yt}}\right)}^2$$ From the previous step, $$\frac{4(x+y+z+t{)}^2}{(x+z{)}^2(y+t{)}^2}\le {\left(\frac{1}{\sqrt{xz}}+\frac{1}{\sqrt{yt}}\right)}^2$$ Therefore, $$\frac{(x+y)(z+t)}{xyzt}\ge {\left(\frac{1}{\sqrt{xz}}+\frac{1}{\sqrt{yt}}\right)}^2\ge \frac{4(x+y+z+t{)}^2}{(x+z{)}^2(y+t{)}^2}$$ which proves the required inequality.