Homework inequality -- Show that (a+1)(b+1)(c+1)(d+1) < 8(abcd+1)

AI Thread Summary
The discussion focuses on proving the inequality (a+1)(b+1)(c+1)(d+1) < 8(abcd+1) for a, b, c, d greater than 1. Initial attempts included applying Cauchy-Schwarz and AM-GM inequalities, but without success. A suggestion was made to use induction and to break the problem down into pairs of variables. Participants noted the importance of recognizing that ab > 1 and cd > 1 to simplify the expressions. The conversation highlights collaborative problem-solving in tackling complex inequalities.
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Homework Statement


For a,b,c,d >1, Show that (a+1)(b+1)(c+1)(d+1) < 8(abcd+1)

Homework Equations


How to show this?

The Attempt at a Solution


I could show for two variables, (a+1)(b+1)<2(ab+1). Tried C-S, AM-GM inequalities in different form and variable transformations. But still no result. It's my daughters test question. Any help is appreciated.
 
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Can this be done by induction?
 
(a+1)(b+1)<2(ab+1)
(d+1)(c+1)<2(dc+1)

(a+1)(b+1)(d+1)(c+1)< 4(ab+1)(dc+1)

Can you complete now ?
 
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Likes Vassili sansford
ssd, Buffu has given a nice lead. Recognize that ##ab>1## and ##cd>1##. So replace these in your expressions for ##a## and ##b##
 
Thanks a lot to both of you.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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