Tukukkk

Given that $frac{1}{a}+frac{1}{b}+frac{1}{c}=1$ show that:$$(a+1)(b+1)(c+1)ge 64$$
My attempt: First I tried expanding the LHS getting that$$abc+ab+bc+ca+a+b+c ge 63$$
I applied Cauchy-Schwarz on $(a+b+c)(frac{1}{a}+frac{1}{b}+frac{1}{c})$ getting that $a+b+cge9$.

Then I also tried to manipulate the first condition and got $abc=ab+bc+ca$, then I applied AM-GM on $a+b+c$ getting the following$$a+b+cge3sqrt[3]{abc}$$
Finally I substituted $ab+bc+ca=abc$ in my initial expression, getting:$$2abc+(a+b+c)ge63$$The last thing I tought about was that I have both $a+b+cge9$ and $a+b+cge3sqrt[3]{abc}$ so if I somehow related them I would have $sqrt[3]{abc} ge 3 rightarrow abcge27$ and with this conditions the problem would follow by summing, but the direction of the inequality is not allowing me to do as intended...

I'm stuck here, have tried lot of other things but nothing really worked, also partial help is appreciated!

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
$$ sqrt[3]{abc} geq frac{3}{frac{1}{a}+frac{1}{b}+frac{1}{c}}=3.$$

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 $$a+b+cgeq 9$$ and we get also $$frac{bc+ac+ab}{3}geq sqrt[3]{(abc)^2}$$ so $$abcgeq 27$$ and $$ab+ac+bcgeq 27$$ putting things together we have
$$(a+1)(b+1)(c+1)=abc+ab+bc+ca+a+b+c+1geq 27+27+10=64$$

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

We need to prove that
$$sum_{cyc}(ln(1+a)-2ln2)geq0$$ or
$$sum_{cyc}left(ln(1+a)-2ln2+frac{9}{4}left(frac{1}{a}-frac{1}{3}right)right)geq0,$$
which is true because
$$f(a)=ln(1+a)-2ln2+frac{9}{4}left(frac{1}{a}-frac{1}{3}right)geq0$$ 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!