Question

If $0<a,b,c<1$ and $a+b+c=2$, then the greatest value of $\frac{a}{1-a}\times \frac{b}{1-b}\times \frac{c}{1-c}$ is

• A. $7$
• B. $4$
• C. $8$
• D. $12$

Answer 1

The correct option is C $8$

Given, $a+b+c=2$

we know that, $A.M.\ge G.M.$

$⟹\frac{a+b+c}{3}\ge \sqrt[3]{3abc}$

$⟹\frac{2}{3}\ge \sqrt[3]{3abc}$

$⟹\sqrt[3]{3abc}\le \frac{2}{3}$

cubing on both sides

$⟹abc\le \frac{8}{27}$ --------------(1)

Similarly

$⟹\frac{1-a+1-b+1-c}{3}\ge \sqrt[3]{3(1-a)(1-b)(1-c)}$

$⟹\frac{3-(a+b+c)}{3}\ge \sqrt[3]{3(1-a)(1-b)(1-c)}$

$⟹\frac{3-2}{3}\ge \sqrt[3]{3(1-a)(1-b)(1-c)}$

$⟹\frac{1}{3}\ge \sqrt[3]{3(1-a)(1-b)(1-c)}$

cubing on both sides

$⟹\frac{1}{(1-a)(1-b)(1-c)}\le 27$ ---------------(2)

multiplying (1) and (2) we get

$⟹\frac{abc}{(1-a)(1-b)(1-c)}\le \frac{8}{27}\ast 27$

$⟹\frac{abc}{(1-a)(1-b)(1-c)}\le 8$

The greatest value of $\frac{abc}{(1-a)(1-b)(1-c)}$ is $8$