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I asked a question here, and also got its generalization:(see tc1729's answer)

$$\prod_{1\le i\le n} (1+a_i)\ge 2^{n}\sqrt{a_1a_2\cdots a_n}=2^n\cdot\sqrt M,$$for$$a_1a_2\cdots a_k = M.$$

But I can't understand that how $$\prod_{1\le i\le n} (1+a_i)$$ attends minimum when $$a_1=\cdots = a_n = \sqrt[n]{M}?$$ Also, as I asked here, $$(1+\sqrt[n]{M})^n\neq2^n\cdot\sqrt M.$$So, if $$\prod_{1\le i\le n} (1+a_i)$$ attends its minimum at $$a_1=\cdots = a_n = \sqrt[n]{M},$$ then min value should be $$(1+\sqrt[n]{M})^n$$ and not $$2^n\cdot\sqrt M.$$

Please help. I am confused.

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Silent
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    See comment on the accepted answer on the other page, mentioning an imprecise formulation. Indeed the mimimum value of $\prod\limits_{i=1}^n(1+a_i)$ is $(1+\sqrt[n]{M})^n$ and $2^n\sqrt{M}$ is only a lower bound of this minimum value. – Did Nov 01 '13 at 09:08
  • Many thanks, Sir. Please just now tell why min. is attended at $$a_1=\cdots = a_n = \sqrt[n]{M}?$$ – Silent Nov 01 '13 at 09:15
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    For example, Lagrange multipliers. – Did Nov 01 '13 at 09:31
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    It also follows from http://math.stackexchange.com/questions/29357/an-inequality-is-it-true-if-it-is-then-how-to-prove-it –  Nov 01 '13 at 15:25

2 Answers2

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To complement Ewan Delanoy's answer, let me include an AM-GM proof of Huygen's as well. We shall prove the homogenized version (i.e. $x_i = b_i/a_i$): $$(a_1^n + b_1^n)(a_2^n + b_2^n) \cdots (a_n^n + b_n^n) \ge (a_1a_2 \cdots a_n + b_1b_2 \cdots b_n)^n$$ for $a_1,\cdots,a_n,b_1,\cdots,b_n > 0$.

If we replace $(a_i,b_i)$ by $(ca_i,cb_i)$ for some positive $c$, we see that it's the same inequality, since the inequality is homogeneous in $a_i,b_i$ on both sides. Scale each $(a_i,b_i)$ so that $a_i^n + b_i^n = 1$, then we need to prove $$1 \ge (a_1a_2\cdots a_n + b_1b_2 \cdots b_n)^n$$ or just $$1 \ge a_1a_2\cdots a_n + b_1b_2 \cdots b_n$$ under the condition $a_i^n + b_i^n = 1$ for each $i$. This clearly follows from AM-GM.

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Putting $x_k=a_k^{\frac{1}{n}}$, one can reformulate the inequality to be shown as

$$ P_n : (1+x_1^n)(1+x_2^n)\ldots (1+x_n^n) \geq (1+x_1x_2\ldots x_n)^n \ (\text{for any nonnegative} \ x_1,x_2,\ldots ,x_n) $$

Let us show $P_n$ by induction on $n$. We have that $P_1$ is an equality, $P_2$ is easily equivalent to $(x_1-x_2)^2 \geq 0$. Suppose now that $n\geq 3$ and that $P_{n-1}$ holds. Let $x_1,x_2, \ldots x_n$ be nonnegative numbers. We want to show that

$$ (1+x_1^n)(1+x_2^n)\ldots (1+x_n^n) \geq (1+x_1x_2\ldots x_n)^n \tag{1} $$

If we put $b=(x_2x_3\ldots x_n)^{\frac{n}{n-1}}$, then applying $P_{n-1}$ to the $n-1$ numbers $x_2,x_3, \ldots x_n$, we see that it suffices to show (1) when all the $x_i$ are equal to $b$. In other words, we may assume without loss of generality that $x_2=x_3=\ldots =x_n$, in which case (1) becomes

$$ (1+x_1^n)(1+x_2^n)^{n-1} \geq (1+x_1x_2^{n-1})^n \tag{2} $$

If we put $y=x_2^n$ and $z=\frac{x_1}{x_2}$, (2) can be rewritten as

$$ (1+yz^n)(1+y)^{n-1} \geq (1+yz)^n \tag{3} $$

So if we put $\delta=(1+yz^n)(1+y)^{n-1} - (1+yz)^n$, we must show that $\delta$ is nonnegative. But expanding the Newton binomials leads to

$$ \delta=\sum_{k=1}^{n-1}n\binom{n-1}{k-1}y^k \bigg(\frac{1-z^k}{k}- \frac{1-z^n}{n}\bigg) \tag{4} $$

So all we need to show now is that the sequence $(\frac{1-z^n}{n})$ is decreasing in $n \geq 1$, for any $z>0$. The algebraic equality

$$ \frac{1-z^{n}}{n}-\frac{1-z^{n+1}}{n+1}= \frac{(z-1)^2}{n(n+1)}\bigg( \sum_{k=0}^{n-1} (k+1)z^k \bigg) \tag{5} $$

finishes the proof.

Ewan Delanoy
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  • +1 for so amazing answer. I think this inequality is called Huygens inequality, right? Will you please explain how using $b$ allows only to work with $x_2=x_3=\ldots =x_n$ ? I will be obliged. – Silent Nov 02 '13 at 02:08
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    Yes, it is called Huygens inequlity. By the induction hypothesis, you have $\prod_{k\geq 2}(1+x_k^n)\geq \prod_{k\geq 2}(1+b^n)$. So to show (1) it suffices to show that $(1+x_1^n)\prod_{k\geq 2}(1+b^n) \geq (1+x_1x_2\ldots x_n)^n$, or in other words $(1+x_1^n)\prod_{k\geq 2}(1+b^n) \geq (1+x_1b^{n-1})^n$. – Ewan Delanoy Nov 02 '13 at 07:07
  • Sir, thanks a lot. Will you please explain $\prod_{k\geq 2}(1+x_k^n)\geq \prod_{k\geq 2}(1+b^n)$? I think applying $P_{n-1}$ goes $\prod_{k\geq 2}(1+x_k^n)\geq (1+b^{n-1})^n$, thus can't derive $\prod_{k\geq 2}(1+x_k^n)\geq \prod_{k\geq 2}(1+b^n)$. – Silent Nov 03 '13 at 01:53
  • @Sush You’re right, I mistyped the definition of $b$ it should be $(x_2x_3\ldots x_n)^{\frac{n-1}{n}}$ : Applying $P_{n-1}$ gives $\prod_{k\geq 2}(1+x_k^n) \geq (1+b)^n$. – Ewan Delanoy Nov 03 '13 at 06:25