Proving that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1\implies (a+1)(b+1)(c+1)\geq 64$ where $a,b,c>0$.

Question

I am trying to solve the following problem:

Let $a,b,c>0$ with $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$. Prove that: $$(a+1)(b+1)(c+1)\geq 64$$

So far, I have gotten that by AM-GM, $(a+1)\geq 2\sqrt{a}$, $(b+1)\geq 2\sqrt{b}$ and $(c+1)\geq 2\sqrt{c}$ so:

$$(a+1)(b+1)(c+1)\geq 8\sqrt{abc} \tag{1}$$

Then using AM-GM on $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, we get that:

\begin{align} \frac{1}{a}+\frac{1}{b}+\frac{1}{c} & \geq 3\times \frac{1}{\sqrt[3]{abc}} \\ 1 & \geq 3\times \frac{1}{\sqrt[3]{abc}} \\ \frac{1}{3} & \geq \frac{1}{\sqrt[3]{abc}} \\ \frac{1}{27} & \geq \frac{1}{abc} \\ abc & > 27 \end{align}

Substituting this into $(1)$, we get:

$$(a+1)(b+1)(c+1)\geq 8\sqrt{27}$$

So I clearly went wrong somewhere, though I'm not sure why. It would be best if you could provide a solution using AM-GM or Muirhead's Inequality. The question seems so simple, where equality happens at $a=b=c=3$, though I can't prove this. Thank you in advance for your solutions and hints!

Answer 1

we use $a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}$,with $x+y+z=1$

we have to prove $(1+x)(1+y)(1+z)\ge 64xyz$

But $$1+x=x+y+z+x\ge 4{(x^2yz)}^{1/4}$$ similarly can you do it for $1+y$ and $1+z$ ?.

Multiplying the above results you get dezired inequality.

Answer 2

Another way.

We can use a homogenization.

We need to prove $$\prod_{cyc}\left(a+\frac{1}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)\geq\frac{64}{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^3}$$ or $$\prod_{cyc}(a^2+ab+ac+bc)\geq64a^2b^2c^2$$ or $$\prod_{cyc}((a+b)(a+c))\geq64a^2b^2c^2$$ or $$\prod_{cyc}(a+b)\geq8abc$$ or $$\prod_{cyc}(a+b)\geq8\prod_{cyc}\sqrt{ab}$$ or $$\prod_{cyc}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)\geq8,$$ which is true by AM-GM.