# What Is The Numerical Value Of 1/A + 1/B + 1/C

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Let $a$, $b$, and $c$ be positive real numbers such that $a+b+c = 1$. Then how to establish the following inequality? $$(1-a)(1-b)(1-c) \geq 8abc.$$

My effort:

Since $a+b+c =1$, we can write $$(1-a)(1-b)(1-c) = 1 - (a+b+c) +(ab+bc+ca) - abc = 1 - 1 +(ab+bc+ca) - abc = (ab+bc+ca) - abc = abc(a^{-1} + b^{-1} + c^{-1} ) - abc = 3abc (\frac{a^{-1} + b^{-1} + c^{-1}}{3}) - abc.$$ What next?

Because $x+y \ge 2\sqrt{xy}$ for any $x>0,y>0$, and $1-a = (a+b+c) - a = b+c$ (similarly for $1-b$ and $1-c$) we have $$(1-a)(1-b)(1-c) = (b+c)(c+a)(a+b) \ge 2\sqrt{bc}\cdot 2\sqrt{ca}\cdot 2\sqrt{ab} = 8abc.$$

You can rewrite $(1-a)(1-b)(1-c)$ as $abc(\frac1a +\frac1b + \frac1c -1)$, meaning that your inequality is equivalent to proving that $$\frac 1a + \frac 1b + \frac 1c \geq 9,$$

an inequality that is simpler to show.

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5xum"s answer here. Using $AM\geq HM$,${\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\over{3}$$\geq$${3}\over{a+b+c}$So, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq9$

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Show that the following inequality is true: $(\frac{1}{a} + \frac{1}{bc}) (\frac{1}{b} + \frac{1}{ca})(\frac{1}{c} + \frac{1}{ab}) \geq 1728$