Since $1=a+b+c,$ we are left to prove that $(2a+3b+c)(a+2b+3c)(3a+b+2c) \geq 216abc,$ which follows by applying the AM-GM inequality.

Because $\frac{1 + a}{b} + 2 = \frac{a + 2b + 1}{b} = \frac{2a + 3b + c}{b}$ we have $$ LHS = \frac{\prod(2a + 3b + c)}{abc} = \frac{\prod(a + a + b + b + b + c)}{abc} \stackrel{AM - GM}{\ge} \frac{\prod (6\sqrt[6]{a^2b^3c})}{abc} = 216. $$Equality holds when $a = b = c = \frac{1}{3}$.

$\left ( \frac{1+a}{b}+2 \right ) \left ( \frac{1+b}{c}+2 \right ) \left ( \frac{1+c}{a}+2 \right ) \geq \left( \sqrt[3]{ \frac{(1+a)(1+b)(1+c)}{abc}}+2\right )^3 \geq \left ( \frac{1}{\sqrt[3]{abc}} +3 \right )^3 \geq \left ( \frac{3}{a+b+c}+3\right )^3=216$

$$\left ( \frac{1+a}{b}+2 \right ) \left ( \frac{1+b}{c}+2 \right ) \left ( \frac{1+c}{a}+2 \right ) \geq24\sqrt{ \frac{3}{abc}}\geq 216$$

$\frac{2b+a+1}{b}\cdot\frac{2c+b+1}{c}\cdot\frac{2a+c+1}{a}=\frac{3b+2a+c}{b}\cdot\frac{3c+2b+a}{c}\cdot\frac{3a+2c+b}{a}=\frac{(a+b)+(a+c)+(a+c)}{a}\cdot\frac{(b+c)+(b+a)+(b+a)}{b}\cdot\frac{(c+a)+(c+b)+(c+b)}{c}\geq\frac{3\sqrt[3]{(a+b)(a+c)(a+c)}}{a}\cdot\frac{3\sqrt[3]{(b+c)(b+a)(b+a)}}{b}\cdot\frac{3\sqrt[3]{(c+a)(c+b)(c+b)}}{c}\geq\frac{6\sqrt[3]{a\cdot c\sqrt[2]{ab}}}{a}\cdot\frac{6\sqrt[3]{b\cdot a\sqrt[2]{bc}}}{b}\cdot\frac{6\sqrt[3]{c\cdot b\sqrt[2]{ca}}}{c}=216\times\frac{abc}{abc}=216$ Equality holds if $a=b=c=\frac{1}{3}$

Let $a$, $b$ and $c$ be positive real numbers such that $a+b+c=1$. Prove the inequality $$ \left ( \frac{1+a}{b}+1 \right ) \left ( \frac{1+b}{c}+1 \right ) \left ( \frac{1+c}{a}+1 \right )\geq 125$$here

Homogenise to get the equivalent inequality $$(2a + 3b + c)(2b + 3c + a)(2c + 3a + b) \ge 216abc$$, true by AM-GM. Equality hold when $a = b = c = \frac 13$.

steppewolf wrote: Let $a$, $b$ and $c$ be positive real numbers such that $a+b+c=1$. Prove the inequality $$ \left ( \frac{1+a}{b}+2 \right ) \left ( \frac{1+b}{c}+2 \right ) \left ( \frac{1+c}{a}+2 \right )\geq 216.$$When does equality hold? Proposed by Anastasija Trajanova $\leftrightarrow \prod (a+2b+3c)\ge 216abc$ $\prod (a+2b+3c)\ge \prod 3\sqrt[3]{6abc}=216abc$