Not only AM-GM 3 (The Other AM-GM Part 5)

$\begin{aligned} S & = \sum_{cyc} \frac {b+c+d+e}a \\ & = \sum_{cyc} \frac {{\color{#3D99F6}s}-a}a & \small \color{#3D99F6} \text{where }s=a+b+c+d+e \\ & = {\color{#3D99F6}\sum_{cyc} \frac sa} - 5 & \small \color{#3D99F6} \text{By AM}\ge \text{HM}: \sum_{cyc} \frac 1a \ge \frac {25}s \text{ (see below)} \\ & \ge {\color{#3D99F6}25} - 5 \\ & \ge \boxed{20} \end{aligned}$

Bonus: Similarly, in general $\displaystyle \sum_{k \ne j}^n \frac {x_k}{x_j} \ge n^2 - n$

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AM-HM inequality

$\small \begin{aligned} \frac 5{\frac 1a+\frac 1b+\frac 1c+\frac 1d+\frac 1e} & \le \frac {a+b+c+d+e}5 \\ \frac 1a+\frac 1b+\frac 1c+\frac 1d+\frac 1e & \le \frac {25}{a+b+c+d+e} \end{aligned}$

By Cauchy-Schwarz Inequality, we have

$\begin{aligned} \left( \left( \dfrac{1}{ \sqrt{a}} \right)^2 + \left( \dfrac{1}{ \sqrt{b}} \right)^2 + \left( \dfrac{1}{ \sqrt{c}} \right)^2 + \left( \dfrac{1}{ \sqrt{d}} \right)^2 + \left( \dfrac{1}{ \sqrt{e}} \right)^2 \right) \left( (\sqrt{a})^2 + (\sqrt{b})^2 + (\sqrt{c})^2 + (\sqrt{d})^2 + (\sqrt{e})^2 \right) & \geq \left( \dfrac{1}{\sqrt{a}} \cdot \sqrt{a} + \dfrac{1}{\sqrt{b}} \cdot \sqrt{b} + \dfrac{1}{\sqrt{c}} \cdot \sqrt{c} + \dfrac{1}{\sqrt{d}} \cdot \sqrt{d} + \dfrac{1}{\sqrt{e}} \cdot \sqrt{e} \right)^2 \\ \left( \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{d} + \dfrac{1}{e} \right) \left( a+b+c+d+e \right) & \geq 5^2 \\ \dfrac{b+c+d+e}{a} + \dfrac{a+c+d+e}{b} + \dfrac{a+b+d+e}{c} + \dfrac{a+b+c+e}{d} + \dfrac{a+b+c+d}{e} + 5 & \geq 25 \\ \dfrac{b+c+d+e}{a} + \dfrac{a+c+d+e}{b} + \dfrac{a+b+d+e}{c} + \dfrac{a+b+c+e}{d} + \dfrac{a+b+c+d}{e} & \geq 20 \end{aligned}$

Hence, The answer is $\boxed{20}$

My ways of solving this problem. Please check out if there are any mistakes! Thanks!