Proof of this simple inequality: $\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \geq 4$

Question

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

Answer 1

You can also do the proof in the following manner. $$\dfrac{a}{b}+\dfrac{b}{c}\ge 2\sqrt{\dfrac{a}{c}}$$$$\dfrac{c}{d}+\dfrac{d}{a}\ge 2\sqrt{\dfrac{c}{a}}$$So, $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}\ge 2\left(\sqrt{\dfrac{a}{c}}+\sqrt{\dfrac{c}{a}}\right)\ge 2\left(2\sqrt{\sqrt{\dfrac{a}{c}}\cdot\sqrt{\dfrac{c}{a}}}\right)=4 $$

Answer 2

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 $$x^4+y^4+z^4+t^4-4xyzt=(x^2-y^2)^2+(z^2-t^2)^2+2(xy-zt)^2\geq0$$