5

Suppose $a_1, a_2,\dots , a_n$ are $n$ positive real numbers with $a_1a_2 \dots a_n = 1$. Then what is the minimum value of $(1 + a_1)(1 + a_2). . .(1 + a_n)$ ?

I think $(1 + a_1)(1 + a_2). . .(1 + a_n)$ takes its minimum value when $a_1=a_2=\dots=a_n=1$ and thus the minimum value is $2^n$.

I don't know how to prove it. Please help.

Martin Sleziak
  • 56,060
Silent
  • 6,668

4 Answers4

11

No need for calculus here; by AM-GM we have $1+a_1\ge2\sqrt{a_1},\ldots, 1+a_{n}\ge 2\sqrt{a_n}$ and multiplying them yields $$\prod_{1\le i\le n} (1+a_i)\ge 2^{n}\sqrt{a_1a_2\cdots a_n}=2^n,$$ with equality when $a_1=\cdots = a_n = 1$. In general if $a_1a_2\cdots a_k = M$ for some positive real $M$, then we can give a lower bound on the minimum value, which is similarly computed to be $2^n\cdot \sqrt{M}$, with equality at $a_1=\cdots = a_n = \sqrt[n]{M}$. The absolute minimum, as discussed below, is $(1+\sqrt[n]{M})^n$.

tc1729
  • 3,217
  • Many many thanks, Sir/Mam! – Silent Nov 01 '13 at 02:25
  • Sir/Mam, many thanks for so quick response. Will you just please answer how $(1+\sqrt[n]{M})^n=2^n\sqrt M$ ? – Silent Nov 01 '13 at 05:40
  • I asked here, and got that $(1+\sqrt[n]{M})^n\neq2^n\sqrt M$. Will you please explain your last sentences more so that I can get it? – Silent Nov 01 '13 at 06:05
  • 1
    I'm just plugging the value of $a_1a_2\cdots a_k=M$ into the RHS of the inequality... – tc1729 Nov 01 '13 at 06:06
  • 2
    Sure but mentioning "the minimum value" leads one to believe that $2^n\sqrt{M}$ is the minimum value of the LHS--which it is not except when $M=1$. – Did Nov 01 '13 at 09:05
4

Use the fact that $$1+a_k \geq 2\sqrt{a_k}$$ to conclude what you want.

2

accepted answer is true only for the case $~a_{1}a_{2}\cdots a_{n}=1$.
for example, if $a_{1}a_{2}\cdots a_{n}=10$, it doesn't work any more.

Huygens inequality : $(1+a_{1})(1+a_{2})\cdots(1+a_{n})\geq (1+\sqrt[n]{a_{1}a_{2}\cdots a_{n}})^{n}$

by Huygens inequality, the minimum value is $2^{n}$

chloe_shi
  • 2,885
  • 2
    Why cite obscure results? My answer works perfectly fine... in the case that $a_1\cdots a_n = 10$, the maximum value is just $2^n\sqrt{10}$, with equality at $a_1=\cdots = a_n = \sqrt[n]{10}$. – tc1729 Nov 01 '13 at 02:58
  • 1
    you used AM-GM and watch the equality condition. – chloe_shi Nov 01 '13 at 04:36
  • @tc1729, will you please explain more why your ans. holds even for $a_1\cdots a_n = 10$ (by editing your answer)? and shouldn't it be minimum value (the $2^n\sqrt{10}$) – Silent Nov 01 '13 at 05:28
0

Let $A=\prod_{k=1}^n (1+a_k)$. (A isn't a constant value)

Just use the fact that you can pair up each term of A so that one term of the pair is a reciprocal of the another. For example, there is such a pair consisting of $x$ and $y$. Since $\prod^{n}_{k=1}a_k=1$, $xy=1$. Using this fact and AM-GM inequality, it can be obtained that:

$A=(1^n+\prod^{n}_{k=1}a_k)+(x+y)+(...)+(...)...$ (Such pairs are summed up within the parenthesis)

Since $x+y\ge2 \sqrt{1}$, $A\ge2 \times $(the number of the parenthesis)

Since the number of terms of A is ${2^n}$, $A\ge2 \times 2^{n-1}=2^n$

Math.StackExchange
  • 3,786