An algebra problem by Aly Ahmed

Claim: The minimum value of the expression 'approaches' $1$ $\text{min}\left (\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}+\frac{1}{1+t}\right)\rightarrow 1$ Proof: The condition $xyzt=1$ motivates us to make the substitution $x=\frac{a}{b} \;,\;\; y=\frac{b}{c} \;,\;\; z=\frac{c}{d} \;,\;\; t=\frac{d}{a}$ So the above expression changes to $\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}+\frac{1}{1+t}=\frac{b}{b+a}+\frac{c}{c+b}+\frac{d}{d+c}+\frac{a}{a+d}$ Let $(a,b,c,d)=\left (k,\sqrt{k},1,\frac{1}{k} \right )$ , then when $k$ becomes very large, $\frac{b}{b+a}+\frac{c}{c+b}+\frac{d}{d+c}+\frac{a}{a+d}=\lim_{k\rightarrow \infty}\left (\frac{\sqrt{k}}{\sqrt{k}+k}+\frac{1}{1+\sqrt{k}}+\frac{1}{k+1}+\frac{k^2}{k^2+1}\right)=1$ So it suffices to show that, for all positive reals $\frac{b}{b+a}+\frac{c}{c+b}+\frac{d}{d+c}+\frac{a}{a+d}\ge 1$ Applying Titu's Lemma/Cauchy-Schwarz inequality , $\sum_{cyc}\frac{b}{b+a}=\sum_{cyc}\frac{b^2}{b^2+ab}\ge\frac{(a+b+c+d)^2}{a^2+b^2+c^2+d^2+ab+bc+cd+da}\ge 1$ Hence, we have shown that the expression never takes values less than $1$ . Therefore, $\text{inf}\left( \frac{b}{b+a}+\frac{c}{c+b}+\frac{d}{d+c}+\frac{a}{a+d}\right)=\text{inf}\left (\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}+\frac{1}{1+t}\right)= 1$