Mathematical Reflections 2016, Issue 1 - Problem S362

Problem:
Let $0 < a,b,c,d \leq 1$. Prove that $$\dfrac{1}{a+b+c+d} \geq \dfrac{1}{4}+\dfrac{64}{27}(1-a)(1-b)(1-c)(1-d).$$

Proposed by An Zhen-ping, Xianyang Normal University, China

Solution:
The given inequality can be written as $$\dfrac{4-a-b-c-d}{4(a+b+c+d)} \geq \dfrac{64}{27}(1-a)(1-b)(1-c)(1-d).$$
By the AM-GM Inequality, we have $$\dfrac{64}{27}(1-a)(1-b)(1-c)(1-d) \leq \dfrac{64}{27}\left(\dfrac{4-a-b-c-d}{4}\right)^4=\dfrac{1}{4}\cdot\dfrac{1}{27}(4-a-b-c-d)^4.$$
So, we have to prove that $$\dfrac{1}{4}\cdot\dfrac{1}{27}(4-a-b-c-d)^4 \leq \dfrac{4-a-b-c-d}{4(a+b+c+d)},$$ i.e. $$\dfrac{1}{27}(4-a-b-c-d)^3 \leq \dfrac{1}{a+b+c+d}.$$ Let $x=a+b+c+d$. We have to prove that $x(4-x)^3 \leq 27$ for any $x \in (0,4]$, but this is true because $$27-x(4-x)^3=(x-1)^2(x^2-10x+27) \geq 0$$ for all $x \in (0,4]$. The equality holds if and only if $x=1$ and $1-a=1-b=1-c=1-d$, i.e. if and only if $a=b=c=d=\dfrac{1}{4}$.