Proof by Induction: If $x_1x_2\dots x_n=1$ then $x_1 + x_2 + \dots + x_n\ge n$

Question

If $x_1,x_2,\dots,x_n$ are positive real numbers and if $x_1x_2\dots x_n=1$ then $x_1 + x_2 + \dots + x_n\ge n$

There is a step in which I am confuse. My proof is as follows (it must be proven using induction).

By induction, for $n=1$ then $x_1=1$ and certainly $x_1\ge1$. Suppose $x_1 + x_2+\dots+x_n\ge n$ (does this mean that $x_1x_2\dots x_n = 1$ also hold?) and $x_1x_2\dots x_n x_{n+1} = 1$ hold. Then

\begin{align} x_1 + x_2 + \dots+x_n+x_{n+1} &\ge n + x_{n+1} \\ &=n+2-1/x_{n+1} \\ &=n+2-x_1x_2\dots x_n. \end{align}

My problem is in the last step. As I wrote before, I don't think $x_1x_2\dots x_n = 1$ should hold, because if this is the case then $x_{n+1}=1$.

EDIT:

In the itermediate step I used that $x + 1/x \ge 2$, where $x>0$.

Answer 1

Hint: By the AM-GM inequality we get

$$\frac{x_1+x_2+...+x_n}{n}\geq \sqrt[n]{x_1\cdot x_2\cdot …\cdot x_n}$$

Answer 2

Your proof is wrong, because you don't know whether $x_1x_2\ldots x_n=1$ and therefore you don't know whether $x_1+x_2+\cdots+x_n\geqslant n$.

Note that your inequality is a particular case os the AM-GM inequality: since the geometric mean of $x_1,x_2\ldots,x_n$ is $1$, its arithmetic mean is at least $1$, and this means that $x_1+x_2+\cdots+x_n\geqslant n$.

Answer 3

An alternative proof to AM-GM inequality is Lagrange Multiplier method.

We are minimizing $$x_1+x_2+...+x_n$$ subject to $$x_1x_2x_3...x_n=1$$

That gives us$$ <1,1,1,...,1>=\lambda<x_2x_3...x_n, x_1x_3....x_1x_2...x_{n-1}>$$

$$x_1=x_2=x_3=...=x_n=1$$ $$x_1+x_2+...+x_n=n$$

Answer 4

Proof by Mathematical Induction:

First note that if $a_1\le1$ and $a_2 \ge1$ then $$(a_2-1)(1-a_1)\ge0 \implies \color{red}{a_1a_2 \le a_1 + a_2 -1}.\tag{1}$$ Now, if $x_1 \cdot x_2 \cdots x_n \cdot x_{n+1} = 1$ then one of $x_i$'s must be less than or equal to $1$ and another one must be greater than or equal to $1$. Without loss of generality, we presume $x_n \le 1$ and $x_{n+1} \ge 1$. Then $$x_1 \cdot x_2 \cdots x_{n-1} \color{red}{x_{n}\cdot x_{n+1}}=1 \implies x_1+x_2+\cdots+x_{n-1}+\color{red}{x_n\cdot x_{n+1}}\ge n.$$ Using $(1)$, we arrive at the desired result.

Answer 5

Here is a way to prove AM-GM using backwards induction and then using this to prove your proposition:

We will start with $n=2$: $$(x_1+x_2)^2/4-x_1x_2=\frac14(x_1^2+2x_1x_2+x_2^2-4x_1x_2)=\frac14(x_1-x_2)^2\ge0\\\implies\frac{x_1+x_2}{2}\ge\sqrt{x_1x_2}$$

Assume it is true for $2^k$ for some $k$ natural number we will show it is true for $2^{k+1}$;

$$\begin{align}&\frac{x_1+x_2+\ldots+x_{2^k}+x_{2^k+1}+\ldots+x_{2^{k+1}}}{2^{k+1}}\\=&\frac{\frac{x_1+x_2+\ldots+x_{2^k}}{2^{k}}+\frac{x_{2^k+1}+\ldots+x_{2^{k+1}}}{2^{k}}}{2}\\\overset{\mbox{assumption}}{\ge}&\frac{({x_1+x_2+\ldots+x_{2^k}})^{2^{-k}}+({x_{2^k+1}+\ldots+x_{2^{k+1}}})^{2^{-k}}}{2}\\\overset{\mbox{base case(n=2)}}{\ge}&\sqrt{({x_1x_2\ldots x_{2^k}})^{2^{-k}}({x_{2^k+1}\ldots x_{2^{k+1}}})^{2^{-k}}}\\=&({x_1x_2\ldots x_{2^k}}{x_{2^k+1}\ldots x_{2^{k+1}}})^{2^{-(k+1)}}\end{align}$$

Now let's prove that if it is true for $n$ it is true for all natural numbers less than $n$, take $n<2^k$ for some $k$:$$\mbox{denote }a=\frac{x_1+x_2+\ldots+x_n}{n}$$Hence$$a=\frac{x_1+x_2+\ldots+x_n+(2^k-n)a}{2^k}=\frac{x_1+x_2+\ldots+x_n+\overbrace{a+\ldots+a}^{2^k-n~times}}{2^k}$$Because we know that the inequality holds for $2^k$ we have$$a\ge(x_1x_2\ldots x_n\overbrace{a\ldots a}^{2^k-n~times})^{2^{-k}}=(x_1x_2\ldots x_n{a}^{2^k-n})^{2^{-k}}\\\implies a^{2^k}\ge x_1x_2\ldots x_n{a}^{2^k-n}\implies a^{n}\ge x_1x_2\ldots x_n\\\implies a=\frac{x_1+x_2+\ldots+x_n}{n}\ge (x_1x_2\ldots x_n)^{\frac1n}$$

---

Therefore we showed that the AM-GM inequality holds if $n=2^k$ and also holds if $n<2^k$ for some $k$, therefore it is true for all $n$.

Now we can put $x_1x_2\dots x_n=1$ and by the inequality we have $$\frac{x_1+x_2+...+x_n}{n}\geq \sqrt[n]{x_1 x_2\ldots x_n}=1\\\implies x_1+x_2+...+x_n\ge n$$

---

Maybe it is a little over the top to do it like this but I like this way.

Answer 6

In fact， we may prove the more stonger one

Let $x_1,x_2,\cdots,x_n$ be positive numbers. Then $$\frac{1}{n}(x_1+x_2+\cdots+x_n)\geq \sqrt[n]{x_1x_2\cdots x_n}.\tag1$$

Put $x_1x_2\cdots x_n=1$ into $(1)$. You will get yours.

Inductive Proof

Obviously, $(1)$ holds for $n=1$, which is trivial.

Notice that, from $(\sqrt{x_1}-\sqrt{x_2})^2 \geq 0$, we may have $\dfrac{1}{2}(x_1+x_2) \geq \sqrt{x_1x_2}$, which shows $(1)$ holds for $n=2$.

Assume that $(1)$ holds for $n=k$, namely, $$\frac{1}{k}(x_1+x_2+\cdots+x_k)\geq \sqrt[k]{x_1x_2\cdots x_k}.$$ When $n=k+1$, denote $\dfrac{1}{k+1}(x_1+x_2+\cdots+x_k+x_{k+1})=x.$ Then, \begin{align*} x&=\frac{1}{2k}[(x_1+x_2+\cdots+x_k)+x_{k+1}+(k-1)x]\\&=\frac{1}{2}\left[\frac{1}{k}(x_1+x_2+\cdots+x_k)+\frac{1}{k}(x_{k+1}+(k-1)x)\right]\\ &\geq \frac{1}{2}\left(\sqrt[k]{x_1x_2\cdots x_k}+\sqrt[k]{x_{k+1}x^{k-1}}\right)\\ &\geq \sqrt[2k]{x_1x_2\cdots x_k \cdots x_{k+1}x^{k-1}}, \end{align*} namely, $$x^{2k} \geq x_1x_2\cdots x_kx_{k+1}x^{k-1},$$ i.e. $$x^{k+1} \geq x_1x_2\cdots x_kx_{k+1}.$$Thus, $$\frac{1}{k+1}(x_1+x_2+\cdots+x_k+x_{k+1})\geq \sqrt[k+1]{x_1x_2\cdots x_kx+{k+1}},$$which shows that $(1)$ holds for $n=k+1.$ By induction, $(1)$ holds for all $n=1,2,\cdots$.