Inequality for positive real numbers less than $1$: $8(abcd+1)>(a+1)(b+1)(c+1)(d+1)$

Question

If $a,b,c,d$ are positive real numbers, each less than 1, prove that the following inequality holds: $$8(abcd+1)>(a+1)(b+1)(c+1)(d+1).$$

I tried using $\text{AM} > \text{GM}$, but I could not prove it.

Answer 1

Given that $0

$\therefore$ $-1

Now \begin{align*} &(a-1)(b-1)>0 \\ \Rightarrow& ab - (a+b) +1 >0 \\ \Rightarrow& (ab+1) > a+b \\ \Rightarrow& (ab+1)+(ab+1) > (ab+1)+(a+b) \\ \Rightarrow& 2(ab+1) > (ab+1)+(a+b) \\ \Rightarrow& 2(ab+1) > (a+1)(b+1) \tag{I} \end{align*}

Similarly $$2(cd+1) > (c+1)(d+1)$$

$\therefore$ $0

Similarly $0

Now, \begin{align*} 4(ab+1)(cd+1) &> (a+1)(b+1)(c+1)(d+1) \\ \Rightarrow (a+1)(b+1)(c+1)(d+1) &< 4(ab+1)(cd+1) \tag{II} \end{align*}

As $(I)$ \begin{align*} 2(abcd+1) &> (ab+1)(cd+1) \\ \Rightarrow (ab+1)(cd+1) &< 2(abcd+1) \tag{III} \end{align*}

Putting these value in $(II)$,we get \begin{align*} (a+1)(b+1)(c+1)(d+1) &< 4\left\{2(abcd+1)\right\} \\ (a+1)(b+1)(c+1)(d+1) &< 8(abcd+1) \end{align*} answered by SAROJ GHOSH .

Answer 2

First, note that $2^0(a+1) \ge a + 1$.

Then we prove by induction that $2^{k-1}(x_1\dots x_k+1) > \prod_{i=1}^k(x_i + 1)$. For the ease of notation, there I will focus on the case $k = 4$.

$$(d+1)(a+1)(b+1)(c+1)\le 4(d+1)(abc+1)< 4(abcd+abc+1+d)< 8(abcd+1)$$

The last inequality is provided by

$$abc+d< abcd+1 \Leftarrow Q+d0,$$ where $0 < Q = abc < 1.$

Answer 3

your inequality is equivalent to $$7\,abcd-abc-abd-acd-bcd-ab-ac-ad-bc-bd-cd-a-b-c-d+7>0$$ combining usefull sums we obtain $$(1-a)(1-bcd)+(1-b)(1-acd)+(1-c)(1-abd)+(1-d)(1-abc)+(1-ad)(1-bc)+(1-bd)(1-ac)+(1-cd)(1-ab)>0$$ since we have $$0

Answer 4

Let $f(a,b,c,d)=8(abcd+1)-(a+1)(b+1)(c+1)(d+1)$.

Hence, $f$ is a linear function of $a$, of $b$, of $c$ and of $d$. Thus, $$\min_{\{a,b,c,d\}\subset[0,1]}f=\min_{\{a,b,c,d\}\subset\{0,1\}}f=f(1,1,1,1)=0$$ and since the minimum does not occur, we are done!