We introduce three new variables $x,y,z$ such that $x=a-1, y=b-1$ and $c=z-1.$ Substituting these values into the LHS, we get $$\sum_{cyc}\frac{(x+1)(y+1)}{z}=\sum_{cyc} \left( \frac{xy}{z}+\frac{x}{z}++\frac{y}{z}+\frac{1}{z}\right).$$By AM-GM, $$\frac{ \frac{xy}{z}+\frac{x}{z}+\frac{y}{z}+\frac{1}{z}+\frac{yz}{x}+\frac{y}{x}+\frac{z}{x}+\frac{1}{x}+\frac{zx}{y}+\frac{z}{y}+\frac{x}{y}+\frac{1}{y}}{12} \ge \sqrt[12]{\frac{x^4 y^4 z^4}{x^4 y^4 z^4}}=1.$$The equality case is when all of the terms in the numerator are equal, which only holds when $x=y=z=1$ which is $a=b=c=2.$ Therefore we have shown that $\sum_{cyc}\frac{(x+1)(y+1)}{z}=\sum_{cyc}\frac{ab}{c-1} \ge 12.$

Much simpler: after setting $x=a-1$, $y=b-1$ and $z=c-1$, we have \[ \frac{(x+1)(y+1)}{z} \ge \frac{2\sqrt{x}\cdot 2\sqrt{y}}{z} = 4\frac{\sqrt{xy}}{z}. \]Now, apply AM-GM to resulting terms.

Let $ a,b,c>1 $ . Prove that $$a^2b^2c^2 \ge 64(a-1)(b-1)(c-1)$$$$ \frac{a^2b^2}{(c-1)^2}+ \frac{2abc^2}{(a - 1)(b -1)} \ge 48$$

parmenides51 wrote: Let $a$, $b$ and $c$ be real numbers larger than $1$. Prove the inequality $$\frac{ab}{c-1}+\frac{bc}{a - 1}+\frac{ca}{b -1} \ge 12.$$When does equality hold? (Karl Czakler) sqing wrote: Let $ a,b,c>1 $ . Prove that $$a^2b^2c^2 \ge 64(a-1)(b-1)(c-1)$$$$ \frac{a^2b^2}{(c-1)^2}+ \frac{2abc^2}{(a - 1)(b -1)} \ge 48$$

Attachments:

By Cauchy: \[ \sum_{cyc} \frac{ab}{c-1} \cdot \sum_{cyc} ab(c-1) \geq \left( \sum_{cyc} ab \right)^2 \]It suffices to show: \[ \frac{\left(\sum_{cyc} ab \right)^2}{\sum_{cyc} ab(c-1)} \geq 12 \]\[ \left(\sum_{cyc} ab \right)^2 \geq 12\cdot 3abc - 12\sum_{cyc} ab \]\[ \left(\sum_{cyc} ab \right)^2 + 12 \sum_{cyc} ab \geq 36 abc \]By AM-GM: \[ \left( \sum_{cyc} ab \right)^2 + 12\sum_{cyc} ab \geq \left( 3\sqrt[3]{(abc)^2} \right)^2 + 12\cdot 3\sqrt[3]{(abc)^2} \]Set $s = (abc)^{\frac{1}{3}} > 1$: \[ 9s^4 + 36s^2 \geq 36s^3 \]\[ 9s^4-36s^3+36s^2 = 9s^2(s-2)^2 \geq 0 \]Thus we have proven the inequality. Equality holds for $a=b=c=s = 2$ because of AM-GM

it's simple. let a=1+x,b=1+y,c=1+z. then, use am-gm.

by using $AM \ge HM$ inequality we get $\frac{(\sum ab)^2}{3abc-(\sum ab)}\ge12$ $(\sum ab)^2 \ge 36abc-12(\sum ab)$ $(\sum ab)^2+12(\sum ab)\ge 36abc$ And by $AM \ge GM$ inequality we get $9(\sqrt[3]{(abc)^2})^2+36 \sqrt[3]{(abc)^2}\ge36abc$ Denote $\sqrt[3]{abc}=t$.Then our inequality become $9t^{4}+36t^{2}\ge36t^{3}$ And that is equivalent to $9t^2(t-2)^2 \ge 0$ And we're done.