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Question: If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$

Source: ISI BMath UGB 2010

My approach: We have $0<a,b,c<1$. Therefore we have $$0>-a,-b.-c>-1\implies 1>1-a,1-b,1-c>0\implies 0<1-a,1-b,1-c<1.$$

Thus by AM-GM inequality we have $$\frac{(1-a)+(1-b)+(1-c)}{3}\ge ((1-a)(1-b)(1-c))^{\frac{1}{3}}\\ \implies \frac{1}{3}\ge ((1-a)(1-b)(1-c))^{\frac{1}{3}}\\ \implies \frac{1}{(1-a)(1-b)(1-c)}\ge 27.$$

Also by AM-GM inequality we have $$\frac{a+b+c}{3}\ge (abc)^\frac{1}{3}\\ \implies \frac{2}{3}\ge (abc)^\frac{1}{3}\\\implies abc\le \frac{8}{27}\\\implies \frac{1}{abc}\ge \frac{27}{8}.$$

This implies that, we have $$\frac{1}{abc(1-a)(1-b)(1-c)}\ge \frac{27^2}{8}\\ \implies \frac{(abc)^2}{abc(1-a)(1-b)(1-c)}\ge \frac{(27abc)^2}{8}\\\implies \frac{abc}{(1-a)(1-b)(1-c)}\ge \frac{(27abc)^2}{8}.$$

How to proceed after this?

tryst with freedom
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Sanket Biswas
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    Your approach won't work because $\frac{(27abc)^2}{8}\leq 8$. But if you write $x=1-a$, $y=1-b$, and $z=1-c$, then $$\frac{abc}{(1-a)(1-b)(1-c)}=\frac{(y+z)(z+x)(x+y)}{xyz},$$ which is $\ge 8$ by AM-GM. – Batominovski Apr 04 '20 at 13:02

3 Answers3

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Hint: The inequality is: $$\frac{a}2\cdot\frac{b}2\cdot\frac{c}2 \geqslant (1-a)\cdot(1-b)\cdot(1-c)$$

which follows from Karamata's inequality as $\left(1-a, 1-b, 1-c\right) \succ \left(\dfrac{a}2, \dfrac{b}2, \dfrac{c}2\right)$, and $t\mapsto \log t$ is concave.

Macavity
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You may set $$u=1-a,\; v= 1-b,\; w = 1-c$$

$$\Rightarrow u+v+w= 1 \text{ and } u,v,w\in (0,1)$$

To show is now

$$(1-u)(1-v)(1-w)\geq 8uvw$$

or, after expanding and using $u+v+w=1$

$$uv+vw+wu \geq 9uvw$$

or, because auf AM-HM

$$\frac 1 3 = \frac{u+v+w}{3}\geq \frac 3{\frac 1w + \frac 1u + \frac 1v}$$ which is true.

trancelocation
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Your inequality is the same as $$\sqrt{(1 - a)(1 - b)}\sqrt{(1 - b)(1 - c)}\sqrt{(1 - a)(1 - c)} \leq {abc \over 8}$$ By the AM-GM inequality you have $$\sqrt{(1 - a)(1 - b)} \leq {1 - a + 1 - b \over 2} = {c \over 2}$$ $$\sqrt{(1 - b)(1 - c)} \leq {1 - b + 1 - c \over 2} = {a \over 2}$$ $$\sqrt{(1 - a)(1 - c)} \leq {1 - a + 1 - c \over 2} = {b \over 2}$$ Multiplying these together gives the desired inequality.

Zarrax
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