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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!

JC12
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  • @Alearner The question asks for a proof that $(a+1)(b+1)(c+1)\geq 64$, though if $(a+1)(b+1)(c+1)\geq 8\sqrt{27}$, then this would be false since there exists $a,b,c$ such that $(a+1)(b+1)(c+1)< 64$. Is there a possibility that the question is wrong? – JC12 Sep 30 '20 at 05:02
  • The inequality you have to prove, is an equality when $a=b=c = 3$. On the other hand, the inequalities $a + 1 \geq 2 \sqrt a$ and so on are equalities at $a=1$ and not elsewhere. Therefore, your step is not giving you the right inequality, because you are not preserving the equality cases. In your working, make sure you use inequalities, which are equalities when $a=b=c= 3$. – Sarvesh Ravichandran Iyer Sep 30 '20 at 05:03
  • @Alearner You cannot prove $k\ge 64$ from $k \ge 8\sqrt {27}$. – player3236 Sep 30 '20 at 05:06

2 Answers2

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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.

Hari Ramakrishnan Sudhakar
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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.

Michael Rozenberg
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  • Hello, I know it's been 4 months since this was last active, but I am confused how you got from the first to the second step of your proof. Could you explain? thanks. – Han Xiong Feb 26 '21 at 19:21