Inequality with Four Numbers

Question

I am trying to prove the following inequality for real numbers $a,b,c,d$ all of which are greater than $1$

$8(abcd+1) > (a+1)(b+1)(c+1)(d+1)$

I tried the following approaches :

Used the AM-GM inequality

Tried to form a polynomial whose roots are a,b,c,d

Tried the trigonometric substitution $a=\sec x_1$....

But still I couldn't get any closer to the answer.

Answer 1

$(a-1)(b-1)>0$ as $a,b>1 \implies ab+1>a+b $, or $2(ab+1)>ab+1+a+b=(a+1)(b+1)$.

Similarly $(c+1)(d+1)<2(cd+1)$.

Now $ab,cd>1$ then by similar method $(ab+1)(cd+1)<2(abcd+1)$ hence $4(ab+1)(cd+1)<8(abcd+1) $ therefore $(a+1)(b+1)(c+1)(d+1)<8(abcd+1)$

Answer 2

Here is a visual proof. It uses an old Greek approach for proving the distributive law. The first step is to notice that $(a+1)(b+1) = ab+a+b+1$ and $(c+1)(d+1)=(cd+c+d+1)$. Now examine this picture, I know that they all look like the same length, but use your imagination!:

From here, the proof is pretty much done. If you need a HINT notice that since $a,b,c,d > 1$ area $abcd$ is greater than all other possible areas. An notice that there are eight ways to "match" terms. E.g. $abc$ is missing $d$, but then we would have $abc+d < abcd$.