$(\frac{a}{a+b})^5+(\frac{b}{b+c})^5+(\frac{c}{c+d})^5+(\frac{d}{d+a})^5=(\frac{1}{1+\frac{b}{a}})^5+(\frac{1}{1+\frac{c}{b}})^5+(\frac{1}{1+\frac{d}{c}})^5+(\frac{1}{1+\frac{a}{d}})^5 \geq \frac{1}{8}$ There's $\boxed{\frac{b}{a}=x ; \frac{c}{b}=y ; \frac{d}{c}=z ; \frac{a}{d}=t }$ Inequality1 equivalent to: $x,y,z,t \in R^+$ and $xyzt=1$ prove that $(\frac{1}{1+x})^5+(\frac{1}{1+y})^5+(\frac{1}{1+z})^5+(\frac{1}{1+t})^5 \geq \frac{1}{8}$ Proof $\frac{1}{2}\cdot(\frac{1}{1+x})^5+\frac{1}{2}\cdot(\frac{1}{1+y})^5+\frac{1}{2^6}+\frac{1}{2^6}+\frac{1}{2^6}\geq 5 \cdot \frac{1}{(1+x)^2}\cdot \frac{1}{2^4}$ (Cauchy's Inequality) (1) $\frac{1}{2}\cdot(\frac{1}{1+y})^5+\frac{1}{2}\cdot(\frac{1}{1+y})^5+\frac{1}{2^6}+\frac{1}{2^6}+\frac{1}{2^6}\geq 5 \cdot \frac{1}{(1+y)^2}\cdot \frac{1}{2^4}$ (Cauchy's Inequality) (2) $\frac{1}{2}\cdot(\frac{1}{1+z})^5+\frac{1}{2}\cdot(\frac{1}{1+z})^5+\frac{1}{2^6}+\frac{1}{2^6}+\frac{1}{2^6}\geq 5 \cdot \frac{1}{(1+z)^2}\cdot \frac{1}{2^4}$ (Cauchy's Inequality) (3) $\frac{1}{2}\cdot(\frac{1}{1+t})^5+\frac{1}{2}\cdot(\frac{1}{1+t})^5+\frac{1}{2^6}+\frac{1}{2^6}+\frac{1}{2^6}\geq 5 \cdot \frac{1}{(1+t)^2}\cdot \frac{1}{2^4}$ (Cauchy's Inequality) (4) (1)+(2)+(3)+(4) $\to$ $(\frac{1}{1+x})^5+(\frac{1}{1+y})^5+(\frac{1}{1+z})^5+(\frac{1}{1+t})^5 \geq \frac{5}{16}\cdot([\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2}]+[\frac{1}{(1+z)^2}+\frac{1}{(1+t)^2}])-\frac{12}{2^6}\geq \frac{5}{16}\cdot (\frac{1}{1+xy}+\frac{1}{1+zt})-\frac{12}{2^6}\geq \frac{5}{16}\cdot(\frac{1}{1+xy}+\frac{1}{1+\frac{1}{xy}})-\frac{12}{2^6}=\frac{5}{16}\cdot 1-\frac{12}{2^6}=\frac{1}{8}$ Done Inequality1 $\Box$ Facility inequality $x,y \in R^+$ such that $\frac{1}{(1+x)^2}+\frac{1}{(1+y)^2} \geq \frac{1}{1+xy}$ it is true.

$ \frac{1}{(1+x)^{2}}+\frac{1}{(1+y)^{2}}\geq\frac{1}{1+xy} $ $(1+xy)(1+y)^2+(1+xy)(1+x)^2\geq (1+x)^2(1+y)^2$ $\to$ $xy((1+x)^2+(1+y)^2)+1\geq (1+x)^2(1+y)^2-(1+x)^2-(1+y)^2+1= ((1+x)^2-1)((1+y)^2-1)$ $\to$ $xy(x^2+y^2) \geq x^2 \cdot y^2-2xy+1$ $\to$ $(xy-1)^2\geq 0$ done

bayasaa_84 wrote: $\frac{1}{2}\cdot(\frac{1}{1+x})^5+\frac{1}{2}\cdot(\frac{1}{1+y})^5+\frac{1}{2^6}+\frac{1}{2^6}+\frac{1}{2^6}\geq 5 \cdot \frac{1}{(1+x)^2}\cdot \frac{1}{2^4}$ (Cauchy's Inequality) (1) I think it should be $ \frac{1}{2}\cdot(\frac{1}{1+x})^5+\frac{1}{2}\cdot(\frac{1}{1+x})^5+\frac{1}{2^6}+\frac{1}{2^6}+\frac{1}{2^6}\geq 5 \cdot \frac{1}{(1+x)^2}\cdot \frac{1}{2^4}$ and it's not CS,it should be AM-GM

$a+b+c+d+e \geq 5 \cdot \sqrt[5]{abcde}$ $a=b=\frac{1}{2}(\frac{1}{1+x})^5$ $c=d=e=\frac{1}{2^6}$ AM-GM

Triller wrote: Let $a,b,c,d$ be positive real numbers. Prove that \[(\frac{a}{a+b})^{5}+(\frac{b}{b+c})^{5}+(\frac{c}{c+d})^{5}+(\frac{d}{d+a})^{5}\ge \frac{1}{8}\] Let be given positive reals $a$, $b$, $c$. Prove that: $$\frac{a^{3}}{\left(a+b\right)^{3}}+\frac{b^{3}}{\left(b+c\right)^{3}}+\frac{c^{3}}{\left(c+a\right)^{3}}\geq \frac{3}{8}$$h h