Question

Show for all $a,b,c\in R$ &$a+b+c=1;$ then $(1-a)(1-b)(1-c)$ $\ge$ $8abc$

Answer 1

For positive numbers, we know that A.M. > G.M. i.e. the arithmetic mean is greater than their geometric mean.

Here, $a,b,c>0$ and $a+b+c=1$

To prove : $(1-a)(1-b)(1-c)\ge 8abc$

L.H.S. $=(1-a)(1-b)(1-c)=(b+c)(a+c)(a+b)$

Using the inequality mentioned above, we get

$\frac{b+c}{2}\ge \sqrt{bc}$

$\frac{a+c}{2}\ge \sqrt{ac}$

$\frac{a+b}{2}\ge \sqrt{ab}$

Multiplying these three inequalities, we get $\frac{(b+c)(a+c)(a+b)}{8}\ge \sqrt{a^2{b}^2{c}^2}\ge abc$

Hence, $(b+c)(a+c)(a+b)\ge 8abc$ i.e. $(1-a)(1-b)(1-c)\ge 8abc$
Hence proved.