parmenides51 wrote: Let $a, b, c, d > 0$ satisfying $abcd = 1$. Prove that $$\frac{1}{a + b + 2}+\frac{1}{b + c + 2}+\frac{1}{c + d + 2}+\frac{1}{d + a + 2} \le 1$$ My solution: $$\frac{1}{a+b+2}+\frac{1}{b+c+2}+\frac{1}{c+d+2}+\frac{1}{d+a+2}$$$$\le\frac{1}{2}(\frac{1}{\sqrt{ab}+1}+\frac{1}{\sqrt{bc}+1}+\frac{1}{\sqrt{cd}+1}+\frac{1}{\sqrt{da}+1})$$$$=\frac{1}{2}(\frac{1}{\sqrt{ab}+1}+\frac{1}{\sqrt{cd}+1})+ \frac{1}{2}(\frac{1}{\sqrt{bc}+1}+\frac{1}{\sqrt{da}+1})=1.$$

parmenides51 wrote: Let $a, b, c, d > 0$ satisfying $abcd = 1$. Prove that $$\frac{1}{a + b + 2}+\frac{1}{b + c + 2}+\frac{1}{c + d + 2}+\frac{1}{d + a + 2} \le 1$$ $a+b+c+d\geq 4\sqrt{abcd}=4$ $2a+2b+2c+2d\geq 8$ $(a+b)+(b+c)+(c+d)+(d+a)\geq (2)+(2)+(2)+(2)$ $$a+b\geq 2\implies a+b+2\geq 2+2\implies \frac{1}{a+b+2}\leq \frac 14$$$$b+c\geq 2\implies b+c+2\geq 2+2\implies \frac{1}{b+c+2}\leq \frac 14$$$$c+d\geq 2\implies c+d+2\geq 2+2\implies \frac{1}{c+d+2}\leq \frac 14$$$$d+a\geq 2\implies d+a+2\geq 2+2\implies \frac{1}{d+a+2}\leq \frac 14$$