If $abc = (1-a)(1-b)(1-c)$ and $0 \le a,b,c \le 1$ then find minimum of expression:

$$\min \{a(1-c)+b(1-a)+c(1-b)\}$$

The given expression can be rewritten as:

$$a+b+c-ab-ac-bc = S_1-S_2 \tag{1}$$

Then from the given condition:

$$\begin{align} abc &= 1-(a+b+c)+ab+ac+bc-abc \\ S_3 &= 1-S_1+S_2-S_3\\ S_1-S_2 &= 1-2S_3 \tag{2} \end{align}$$

Using $AM \ge GM$

$$\frac{1-S_3 -S_3}{3} \ge\sqrt[3]{S_3^2}$$

for extreme value I set them equal to get:

$$\begin{align} \frac{(1-2S_3)^3}{27}-S^2_3 &= 0 \\ 8S_3^3 +15S_3^2 +6S_3-1 &=0 \\ (S_3 +1)(8S_3^2+7S_3-1) &= 0\\ (S_3+1)^2\left(S_3-\frac{1}{8}\right) &= 0 \end{align}$$

From here we get for minimum value, $S_3 = \frac{1}{8}$. We ignore $S_3 = -1$ because all $a,b,c \ge 0$.

So we can say:

$$\min \{1-2S_3\} = \frac{3}{4}$$

Q. Is there error in any argument (highly probable)

Q. Please provide a better method for inequality, if possible!


I hope you understand my terminology, cannot put it in a better way:

$$S_i = \sum_{cyc} a_1 a_2 a_3 ... ('i'\text{ factors})$$

  • $\begingroup$ AM-GM holds only for non-negative values $\endgroup$ – Tushant Mittal Nov 13 '17 at 10:50
  • $\begingroup$ Yeah so this is wrong anyway.. $\endgroup$ – SJ. Nov 13 '17 at 10:59

We'll prove that $\frac{3}{4}$ is a minimal value.

Let $c=0$.

Hence, $(1-a)(1-b)=0$ and $\sum_{cyc}a(1-c)=a+b-ab=1>\frac{3}{4}.$

Now, let $abc\neq0$, $\frac{a}{1-a}=x$, $\frac{b}{1-b}=y$ and $\frac{c}{1-c}=z$.

Thus, $xyz=1$, $a=\frac{x}{1+x},$ $b=\frac{y}{1+y}$, $c=\frac{z}{1+z}$ and we need to prove that $$\sum_{cyc}\frac{x}{1+x}\left(1-\frac{z}{1+z}\right)\geq\frac{3}{4}$$ or $$\sum_{cyc}\frac{x}{(1+x)(1+z)}\geq\frac{3}{4}$$ or $$4\sum_{cyc}x(1+y)\geq3\prod_{cyc}(1+x)$$ or $$4(x+y+z)+4(xy+xz+yz)\geq3+3(x+y+z)+3(xy+xz+yz)+3xyz$$ or $$x+y+z+xy+xz+yz\geq6,$$ which is true by AM-GM.

The equality occurs for $x=y=z=1$, which says that $\frac{3}{4}$ is the answer.

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  • $\begingroup$ How did you get last inequality? Thanks for answer! $\endgroup$ – SJ. Nov 13 '17 at 11:42
  • $\begingroup$ @samjoe I added something. See now. $\endgroup$ – Michael Rozenberg Nov 13 '17 at 12:32
  • $\begingroup$ Thanks a lot! Do you see any fault in my answer apart from the big blunder? $\endgroup$ – SJ. Nov 14 '17 at 2:07
  • $\begingroup$ @samjoe AM-GM works with non-negative numbers. $\endgroup$ – Michael Rozenberg Nov 14 '17 at 2:41

$$a(1-c)+b(1-a)+c(1-b)=1-2abc\ge \dfrac{3}{4}$$

$$(abc)^2=(a(1-a))\cdot (b(1-b))\cdot(c(1-c))\le \dfrac{1}{4^3}$$

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  • $\begingroup$ Could you please explain what you have done? Thanks! $\endgroup$ – SJ. Nov 13 '17 at 11:13

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