Mathematics - Quadratic Equation Question with Solution | TestHub

Consider $1+a=1+(1-b-c)=(1-b)+(1-c)$.

By AM-GM inequality, for positive numbers $x,y$, we have $\frac{x+y}{2}\ge \sqrt{xy}$, so $x+y\ge 2\sqrt{xy}$.

Thus, $1+a=(1-b)+(1-c)\ge 2\sqrt{(1-b)(1-c)}$.

Similarly, $1+b=(1-a)+(1-c)\ge 2\sqrt{(1-a)(1-c)}$.

And $1+c=(1-a)+(1-b)\ge 2\sqrt{(1-a)(1-b)}$.

Multiplying these three inequalities:

$(1+a)(1+b)(1+c)\ge 8(1-a)(1-b)(1-c)$