Proving a three variables inequality

I think you mean $$a,b,c>0$$ in this case we have $$\frac{a+b+c}{3}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$ so $$a+b+c\geq 9$$ and we get also $$\frac{bc+ac+ab}{3}\geq \sqrt[3]{(abc)^2}$$ so $$abc\geq 27$$ and $$ab+ac+bc\geq 27$$ putting things together we have $$(a+1)(b+1)(c+1)=abc+ab+bc+ca+a+b+c+1\geq 27+27+10=64$$

By the super-additivity of the geometric mean / convexity of $\log(e^x+1)$ $$ (1+a)(1+b)(1+c) \geq (1+\sqrt[3]{abc})^3 $$ and by the GM-HM inequality $$ \sqrt[3]{abc} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=3.$$

Since $ab+bc+ca= abc$ and $$ab+bc+ca\geq 3\sqrt[3]{a^2b^2c^2}\implies abc\geq 27$$ Now $$(a+1)(b+1)(c+1)=2abc+a+b+c+1\geq 55+3\sqrt[3]{27} = 64$$

For positive variables we need to prove that $$\ln\left((1+a)(1+b)(1+c)\right)\geq\ln64$$ or $$\sum_{cyc}(\ln(1+a)-2\ln2)\geq0$$ or $$\sum_{cyc}\left(\ln(1+a)-2\ln2+\frac{9}{4}\left(\frac{1}{a}-\frac{1}{3}\right)\right)\geq0,$$ which is true because $$f(a)=\ln(1+a)-2\ln2+\frac{9}{4}\left(\frac{1}{a}-\frac{1}{3}\right)\geq0$$ for all $a>0$.

Indeed, $$f'(a)=\frac{(a-3)(4a+3)}{4a^2(a+1)},$$ which gives $a_{min}=3$ and since $f(3)=0$, we are done!