Proving inequality by induction only

Question

Suppose $a_1,\dots,a_n$ are real positive numbers s.t. $\prod_{i=1}^n a_i=1$. My book claims that by induction only (i.e. the use of AM-GM is forbidden), one can prove that $$\sum a_i\ge n$$ and that equiality exists if and only if $\forall i,a_i=1$. I tried to prove it:

for $n=2,$ denote $a_1=a,a_2=\frac 1 a$ then: $$0\le \frac{(a-1)^2}{a}=a+\frac{1}{a}-2\Rightarrow a_1+a_2\ge 2.$$

Now we need to prove it for $n+1$. Denote by $b_n=a_na_{n+1}$, By hypothesis $$a_1\cdots a_{n-1} b_n=1\Rightarrow a_1+\dots a_{n-1}+b_n\ge n$$ which means $$a_1+\dots+a_{n-1}+a_n +a_{n+1} \ge n-b_n+a_n+a_{n+1}.$$ That means I need to prove $$a_n+a_{n+1}-a_na_{n+1}\ge 1$$ but I don't know how.

How can I complete the proof?

Answer 1

The inequality you need to prove (I think you should change the $n-1$ to $n+1$ if you follow what you have on the previous line), $$ a_n+a_{n+1}-a_na_{n+1}\geq 1 $$ is not in general true. Take $a_n=a_{n+1}=2$ (which is perfectly valid). This gives zero in the left-hand side.

So, maybe what you can do, is to assume that you, in your product $b_n=a_na_{n+1}$ choose the elements $a_n$ and $a_{n+1}$ that has a certain property?

Answer 2

What we want to show is that $\prod_{i=1}^n a_i=1 \implies \sum a_i\ge n $.

This is certainly true for $n=1$.

Suppose it is true for $n$.

Suppose we have $\prod_{i=1}^{n+1} a_i=1 $. We want to show that $ \sum_{i=1}^{n+1} a_i\ge n+1 $.

As a first try, let's look at $\prod_{i=1}^{n} a_i $. If $A =(\prod_{i=1}^{n} a_i)^{1/n} $ and $b_i =a_i/A $, then $\prod_{i=1}^{n} b_i = 1 $, so $\sum_{i=1}^{n} b_i \ge n $, or $\sum_{i=1}^{n} a_i \ge An $. Therefore $\sum_{i=1}^{n+1} a_i \ge An+a_{n+1} $.

If we can show that $ An+a_{n+1} \ge n+1 $, we are done.

This is the same as $\frac{An+a_{n+1}}{n+1} \ge 1 $. But $1 =\prod_{i=1}^{n+1} a_i =a_{n+1}\prod_{i=1}^{n} a_i =a_{n+1}A^n $, so this is the same as $\frac{An+a_{n+1}}{n+1} \ge a_{n+1}A^n $ or $An+a_{n+1} \ge (n+1)a_{n+1}A^n $ or $An+1/A^n \ge (n+1)a_{n+1}A^n $

By the AM-GM means inequality, since $nA$ is $n$ copies of $A$,

$\begin{array}\\ \frac{ An+a_{n+1}}{n+1} &\ge \sqrt[n]{A^n a_{n+1}}\\ &= \sqrt[n]{(\prod_{i=1}^n a_i) a_{n+1}}\\ &= \sqrt[n]{\prod_{i=1}^{n+1} a_i}\\ &= 1\\ \end{array} $

and, to my pleasant surprise, we are done.

(added even later)

We are not allowed to use the AM-GM inequality. However, I will now show, based on some earlier work of my own, that this particular use on the AM-GM inequality, in which all but one of the values are equal, is a consequence of Bernoulli's inequality.

Bernoulli's inequality, which is easily proved by induction, states that $(1+x)^m \ge 1+mx $ if $x \ge 0$ and $m$ is a positive integer with equality if and only if $m = 1$ or $x = 0$.

Therefore $(1+\frac{v/u-1}{m})^m \ge 1+m\frac{v/u-1}{m} = \frac{v}{u} $ with equality only if $m=1$ or $v/u-1 = 0 $, which is the same as $v = u$. Multiplying by $u^m$, this becomes

$\begin{array}\\ u^{m-1}v &\le u^m(1+\frac{v/u-1}{m})^m \\ &= (u+\frac{v-u}{m})^m \\ &= (\frac{v+(m-1)u}{m})^m \\ \text{or}\\ (u^{m-1}v)^{1/m} &\le \frac{v+(m-1)u}{m}\\ \end{array} $

But this is just the AM-GM inequality with all but one of the values equal!

Therefore, this use of the AM-GM inequality is OK to use, since it is a consequence of Bernoulli's inequality, not the full AM-GM inequality.

(added later)

If there is equality, then $A = a_{n+1} $. Since $1 =A^n a_{n+1} $, $1 = a^{n+1}_{n+1} $, so $a_{n+1} = 1$. This implies that there is equality for $n$, so that $a_n = 1$. This allows us to work our way back down in an odd kind of inverse induction to show that equality at any step shows that at that step and all preceding steps $a_i = 1$.

This is weird, but I think that it is correct.

Answer 3

$$a_n+a_{n+1}-a_na_{n+1}\ge 1 \Leftrightarrow 1 - a_n -a_{n+1} + a_na_{n+1} \le 0. $$

Factorize the LHS and you will see which $a_n$ and $a_{n+1}$ you should pick.