parmenides51 wrote: Let $a, b, c, d \in [0, 1]$. Prove that $$\frac{a}{1 + b}+\frac{b}{1 + c}+\frac{c}{1 + d}+\frac{d}{1 + a}+ abcd \le 3.$$ https://artofproblemsolving.com/community/c6h1434743p8119658 Let $a, b, c, d \in [\frac{1}{2}, 2]$ and $ abcd=1 $. Prove that $$(a+\frac{1}{b})(b+\frac{1}{c})(c+\frac{1}{d})(d+\frac{1}{a})\leq 25$$\begin{align*} P(a,b,c,d)&=\left( ab+\frac ac+1+\frac1{bc} \right)\left( cd+\frac ca+1+\frac1{da} \right) \\ & =\left( a(b+d)+\frac{a+c}c \right)\left( c(b+d)+\frac{a+c}a \right) \\ & =ac(b+d)^2+2(a+c)(b+d)+\frac{(a+c)^2}{ac} \\ & =\frac{(b+d)^2}{bd}+\frac{2(a+c)(b+d)}{\sqrt{abcd}}+\frac{(a+c)^2}{ac} \\ & =\left( \frac{b+d}{\sqrt{bd}}+\frac{a+c}{\sqrt{ac}} \right)^2 \\ & =\left( \sqrt{\frac ac}+\sqrt{\frac ca}+\sqrt{\frac bd}+\sqrt{\frac db} \right)^2 \\ & \leqslant 25. \end{align*}