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

Question

Let $a,b,c>0$ such that $a+b+c=1$. Prove that

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

I tried to plug in $a=1-b-c$ etc. into the inequality, but it doesn't really work. I'm not sure how to even start this.

Answer 1

Note that $(1+a)=(2a+b+c)=((a+b)+(a+c))$. This way, using the inequality $x+y \geq 2\sqrt{xy}$, we get $$ 1+a = (a+b)+(a+c)\geq 2 \sqrt{a+b}\cdot \sqrt{a+c}. $$ We get similar inequalities for $1+b$ and $1+c$. Multiply them to get $$ (1+a)(1+b)(1+c) \geq 8 (a+b)(a+c)(b+c)=8(1-c)(1-b)(1-a), $$ as required.

Answer 2

Hint: use Jensen's inequality with $f(x):=\ln(1+x)-\ln(1-x)$ so $f^{\prime\prime}=\frac{1}{(1-x)^2}-\frac{1}{(1+x)^2}>0$.