sqing wrote: Let $a, b, c$ be the positive real numbers such that $a+b+c+2=abc .$ Prove that $$(a+1)(b+1)(c+1)\geq 27.$$ Proof of Zhangyunhua: $$abc=a+b+c+2\geq 4\sqrt[4]{2abc} \implies abc\geq 8,$$$$ a+1=\frac{a}{2}+\frac{a}{2}+1\geq 3\sqrt[3]{\frac{a^2}{4}} \implies (a+1)(b+1)(c+1)\geq 27\sqrt[3]{\frac{(abc)^2}{64}}=27.$$

sqing wrote: Let $a, b, c$ be the positive real numbers such that $a+b+c+2=abc .$ Prove that $$(a+1)(b+1)(c+1)\geq 27.$$ Generalization of Austria 2019: Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive numbers such that $a_1+a_2+\cdots+a_n+2^n-2n=a_1a_2\cdots a_n.$ Prove that $$(a_1+1)(a_2+1)\cdots (a_n+1)\geq 3^n.$$

sqing wrote: Let $a, b, c$ be the positive real numbers such that $a+b+c+2=abc .$ Prove that $$(a+1)(b+1)(c+1)\geq 27.$$ Proof : $$a+b+c+2=abc \iff \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=1,$$$$1=\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq \frac{3}{\sqrt[3]{(a+1)(b+1)(c+1)}}\implies (a+1)(b+1)(c+1)\geq 27.$$ Let $a, b, c$ be the positive real numbers such that $a+b+c+2=abc .$ Prove that $$(a+1)(b+1)(c+1)\geq(\sqrt{bc}+1)(\sqrt{ca}+1)(\sqrt{ab}+1)\geq 27.$$

We can factorize inequality to get $abc+ab+bc+ca+a+b+c+1 \geq 27$ and now we will use given facts to get $2a+2b+2c+ab+bc+ca \geq 24$, which is enough to proove. From $AM-GM$ inequality we have that $2a+2b+2c+ab+bc+ca \geq 6\sqrt[6]{8a^3b^3c^3}=6\sqrt{2abc}$ and $abc=a+b+c+2 \geq 4\sqrt[4]{2abc}$ or $a^4b^4c^4 \geq 2^9 abc$ or $abc \geq 8$. Now $2a+2b+2c+ab+bc+ca \geq 6\sqrt{2abc} \geq 6\sqrt{16}=24$. And finally, $(a+1)(b+1)(c+1) \geq 27$.

$$a= \frac{y+z}{x}, b = \frac{z+x}{y}, c = \frac{x+y}{z}$$$$\boxed{Tugadi!!!}$$