# Inequality with Four Numbers

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.

$(a-1)(b-1)>0$ 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 $(a+1)(b+1)(c+1)(d+1)<8(abcd+1)$

• this is almost unreadable why dont you use line breaks and punctuation not to mention latex but at least its correct well done – TonyK Aug 27 '15 at 18:02
• I'll try from the next time thanks – Rupsa Aug 27 '15 at 18:05
• @Rupsa What was your thought process behind this elegant solution? – Irrational 3.14 Aug 28 '15 at 1:43
• @Irrational3.14 the solution is based on elementary theorems of inequality – Rupsa Aug 28 '15 at 12:15

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 $abc+d < abcd$.

Let $a=1+x$, $b=1+y$, $c=1+z$ and $d=1+s$.

Hence, $$8(abcd+1)-(1+a)(1+b)(1+c)(1+d)=$$ $$=7xyzt+6(xyz+xys+xzs+yzs)+7(xy+xz+yz+xs+ys+zs)>0$$