Given that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$.Show that:$(a+1)(b+1)(c+1)\ge 64$

Question

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 $a+b+c\ge9$.

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$$a+b+c\ge3\sqrt[3]{abc}$$ 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 $a+b+c\ge9$ and $a+b+c\ge3\sqrt[3]{abc}$ so if I somehow related them I would have $\sqrt[3]{abc} \ge 3 \rightarrow abc\ge27$ 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!

Answer 1

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

Answer 2

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

Answer 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 $$a+b+c\geq 9$$ 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 $$(a+1)(b+1)(c+1)=abc+ab+bc+ca+a+b+c+1\geq 27+27+10=64$$

Answer 4

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!