<s>[INST] Proof of this simple inequality: $\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \geq 4$ Let $a, b, c, d \in \mathbb{R}_{>0}$, then prove that
$\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \geq 4$
Can this be done without using AM-GM inequality, or without using any identity/theorem of inequality? I don't want it to be concise or elegant, I just want rigorous steps that show explicitly how we achieve that result. 
Thanks in advance
 [/INST] Let $\frac{a}{b}=x^4$, $\frac{b}{c}=y^4$, $\frac{c}{d}=z^4$ and $\frac{d}{a}=t^4$ for positives $x$, $y$, $z$ and $t$.
Hence, we need to prove that $x^4+y^4+z^4+t^4\geq4xyzt$, which is true because
$$x^4+y^4+z^4+t^4-4xyzt=(x^2-y^2)^2+(z^2-t^2)^2+2(xy-zt)^2\geq0$$
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