Question

If $a+b+c=1$ and $a,b,c\in R^{+}$ such that $(1-a)(1-b)(1-c)\ge \lambda bac$, then $\lambda =$

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is D $8$

Given, $a+b+c=1$

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

$⟹\frac{a+b+c}{3}\ge \sqrt[3]{3abc}$

$⟹\frac{1}{3}\ge \sqrt[3]{3abc}$

$⟹\sqrt[3]{3abc}\le \frac{1}{3}$

cubing on both sides

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

Similarly

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

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

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

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

cubing on both sides

$⟹\frac{1}{(1-a)(1-b)(1-c)}\le \frac{27}{8}$ ---------------(2)

multiplying (1) and (2) we get

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

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

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

$⟹(1-a)(1-b)(1-c)\ge 8abc$

Therefore $\lambda$ is $8$