- #1

CGandC

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**Summary::**x

Hey, I'm learning calculus and had to prove the following Lemma which is used to prove AM-GM inequality, I had tried to prove it on my own and it is quite different from what is written in my lecture notes.

I have a feeling that my proof of the Lemma is incorrect, but I just don't understand why, can anyone please help? Thanks for any assistance in advance.

**Lemma:**

If ## a_1,a_2,...,a_n ## are real and positive numbers satisfying ## a_1 \cdot a_2 \cdot \cdot \cdot \cdot a_n = 1 ## .

Then prove that ## \sum_{k=1}^n a_k \geq n ## , for any natural number ## n ## .

**My proof:**

Let ## n ## be arbitrary natural number. Let ## a_1,a_2,...,a_n ## be real and positive numbers. I'll prove by induction on ## n ##.

**Base case:**## n=1 ## the Lemma obviously follows.

**Induction step:**Suppose that ## a_1 \cdot a_2 \cdot \cdot \cdot \cdot a_n = 1 ## implies ## \sum_{k=1}^n a_k \geq n ##.

So we'll prove that ## a_1 \cdot a_2 \cdot \cdot \cdot \cdot a_{n+1} = 1 ## implies ## \sum_{k=1}^{n+1} a_k \geq n+1 ##.

Suppose ## a_1 \cdot a_2 \cdot \cdot \cdot \cdot a_{n+1} = 1 ##.

Without loss of generality, assume ## a_n \leq 1 ## and ## a_{n+1} \geq 1 ##.

it follows that:

## a_1 + a_2 + ... + a_n + a_{n+1} = \sum_{k=1}^n a_k + a_{n+1} ##

using the assumption ## \sum_{k=1}^n a_k \geq n ## and ## a_{n+1} \geq 1 ## then ## \sum_{k=1}^n a_k +a_{n+1} \geq n+1 ## so it follows that ## a_1 + a_2 + ... + a_n + a_{n+1} = \sum_{k=1}^n a_k + a_{n+1} \geq n+1 ##. QED

**Proof from notes:**

Base case:## n=1 ## the Lemma obviously follows.

Base case:

**Induction step:**Suppose that ## a_1 \cdot a_2 \cdot \cdot \cdot \cdot a_n = 1 ## implies ## \sum_{k=1}^n a_k \geq n ##.

So we'll prove that ## a_1 \cdot a_2 \cdot \cdot \cdot \cdot a_{n+1} = 1 ## implies ## \sum_{k=1}^{n+1} a_k \geq n+1 ##.

Suppose ## a_1 \cdot a_2 \cdot \cdot \cdot \cdot a_{n+1} = 1 ##.

Without loss of generality, assume ## a_n \leq 1 ## and ## a_{n+1} \geq 1 ##.

it follows that:

## a_1 + a_2 + ... + a_n + a_{n+1} = a_1 + a_2 + ... + a_{n-1} + a_n \cdot a_{n+1} + ( a_n + a_{n+1} - a_n \cdot a_{n+1} ) \geq n + ( a_n + a_{n+1} - a_n \cdot a_{n+1} ) ##

It is left to prove that ## a_n + a_{n+1} - a_n \cdot a_{n+1} ) \geq 1 ##:

## a_n + a_{n+1} - a_n \cdot a_{n+1} = a_{n+1}(1-a_n) - (1-a_n) = ( a_{n+1} -1 )( 1 - a_n ) \geq 0 ## . QED.