Alternate methods to prove $(1+a)(1+b)(1+c)(1+d) \geq 16$ if $abcd =1$.

Question

I found this question some time ago in an Elementary Olympiad book:

If $a, b, c, d$ are positive integers such that $abcd =1$, then prove that $(1+a)(1+b)(1+c)(1+d) \geq 16$.

Evidently this was a direct consequence of Hölder's inequality, so I merely gave the following proof.

$$(1^4+(a^{\frac{1}{4}})^4)(1^4+(b^{\frac{1}{4}})^4)(1^4+(c^{\frac{1}{4}})^4)(1^4+(d^{\frac{1}{4}})^4) \geq (1+(abcd)^{\frac{1}{4}})^4 $$ $$(1+a)(1+b)(1+c)(1+d) \geq (1+1)^4$$ $$(1+a)(1+b)(1+c)(1+d) \geq 16$$

QED.

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However, I was requested by my instructor to try the problem again using only the AM-GM inequality or the Cauchy-Schwarz inequality, because it is supposedly solvable using only those two.

At this point I expanded the expression believing that all the terms so formed would be of the degree $4$, and would, when taken together in groups, yield a power of $abcd$ on the GM side, which could then be added.

Doing so, however, I only get the weaker result that $(1+a)(1+b)(1+c)(1+d) \geq 14$.

I would appreciate a hint for this tantalizing problem.

Answer 1

Hint: $1+x \ge 2\sqrt{x}$. Do this to each variable , and multiply !

Answer 2

By AM-GM, $1+a\ge 2\sqrt{a}.$ So $$(1+a)(1+b)(1+c)(1+d)\ge 16 \sqrt{abcd}=16$$

Answer 3

I guess the following is the easiest one:

Via the AM-GM inequality, $$1+a\geq 2\sqrt a$$ Similarly, obtain for $b,c,d$ and multiply: $$(1+a)(1+b)(1+c)(1+d)\geq 16 \sqrt{abcd}$$ Substitute $abcd=1$ and you are done.

Answer 4

Maybe it means the following way.

By AM-GM $$\prod_{cyc}(1+a)=$$ $$=1+(a+b+c+d)+(ab+ac+ad+bc+bd+cd)+$$ $$+(abc+abd+acd+bcd)+abcd\geq$$ $$\geq1+4+6+4+1=16.$$