Help proving the AM-GM inequality by myself using simple induction.

Question

I am new to this forum. I am currently trying to prove the AM-GM inequality using simple induction on my own using my own method. Most of the proofs I have seen use stronger inequalities or Cauchy's induction. I have read many of the threads related to the proof of this inequality but I was unable to find one that matched my attempt. Here is my attempt:

Theorem. (AM-GM inequality) $$\frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \dots a_n}\, $$ where $(a_i)_{i=0}^{n} \in \mathbb{R}^{+}\, \, \forall i \in \mathbb{N}$.

Proof. We prove the claim by induction on $n$.

Base case. $n = 1$. We get $a_1 \geq a_1$, which is true. $\blacksquare$

Inductive hypothesis. $$\frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \dots a_n}$$

Inductive claim. $$\frac{a_1 + a_2 + \dots + a_n + a_{n+1}}{n+1} \geq \sqrt[n+1]{a_1 a_2 \dots a_n a_{n+1}}$$

Proof of inductive claim.

From the inductive hypothesis, we get $$\frac{a_1 + a_2 + \dots + a_n}{n+1} \geq \frac{n \sqrt[n]{a_1 a_2 \dots a_n}}{n+1}.$$ Adding $\frac{a_{n+1}}{n+1}$ on both sides we obtain $$\frac{a_1 + a_2 + \dots + a_n + a_{n+1}}{n+1} \geq \frac{n \sqrt[n]{a_1 a_2 \dots a_n} + a_{n+1}}{n+1} .$$

Now we claim that $$\frac{n \sqrt[n]{a_1 a_2 \dots a_n} + a_{n+1}}{n+1} \geq \sqrt[n+1]{a_1 a_2 \dots a_n a_{n+1}}$$

Before we prove the above claim, we prove some preliminary lemmas. For simplicity let $$p = \sqrt[n]{a_1 a_2 \dots a_n}$$ and $$q = \sqrt[n+1]{a_1 a_2 \dots a_n a_{n+1}}.$$

Lemma 1. $p > q \iff p > a_{n+1}$.

Proof. $\frac{p}{q} = \frac{\sqrt[n]{a_1 a_2 \dots a_n}}{\sqrt[n+1]{a_1 a_2 \dots a_n} \sqrt[n+1]{a_{n+1}}}$. Hence $$(a_1 a_2 \dots a_n)^{\frac{1}{n} - \frac{1}{n+1}} = (a_1 a_2 \dots a_n)^{\frac{1}{n(n+1)}} > (a_{n+1})^{\frac{1}{n+1}}.$$ Exponentiating both sides by $n+1$, we get $p > a_{n+1}$. By working backwards we can establish the other direction. Similarly, $p < q \iff p < a_{n+1}$ and $p = q \iff p = a_{n+1}$. $\blacksquare$

Lemma 2. $p\sqrt[n]{a_{n+1}} \geq q \iff q \geq 1$.

Proof. We have $(a_1 + a_2 + \dots + a_n + a_{n+1})^\frac{1}{n} \geq (a_1 + a_2 + \dots + a_n + a_{n+1})^\frac{1}{n+1}$ and hence, $$(a_1 + a_2 + \dots + a_n + a_{n+1})^\frac{1}{n(n+1)} \geq 1$$ which implies $$((a_1 + a_2 + \dots + a_n + a_{n+1})^\frac{1}{n+1})^\frac{1}{n} = q^{1/n} \geq 1.$$ Hence we conclude that $q \geq 1$. We can establish the other direction by working backwards. Similarly, $p\sqrt[n]{a_{n+1}} \leq q \iff q \leq 1$. $\blacksquare$

Claim 1. $np + a_{n+1} \geq (n+1)q$

To prove my claim, I considered splitting into 3 cases, by the trichotomy law for positive reals:

1. $a_{n+1} > p$
2. $a_{n+1} < p$
3. $a_{n+1} = p$.

Case 3.

Proof. From lemma 1, we have $p=q$. Hence $(n+1)p = (n+1)q$, which implies the claim. $\blacksquare$

I am confused on how to prove either of case 1 or 2, any hints are welcome. For case 2, I tried writing $p=a_{n+1}+r$, where $r\in \mathbb{R}^{+}$. But I am unable to proceed further. Maybe I am missing an important lemma/observation or I am unable to connect the dots? Or is my proof wrong? I thank you for your time. Again, as I am new to this forum, any feedback would be highly valued and excuse my English if it is wrong.

Edit: The suggested proof only holds for numbers in the form of $2^k$. I have clarified in the question that my method is different.

Answer 1

Claim $1$ can be rewrtitten as $$\left({np+a_{n+1}\over n+1}\right )^{n+1}\ge p^na_{n+1} $$ Divide both sides by $p^{n+1}$ to get an equivalent form $$ \left ({n+p^{-1}a_{n+1}\over n+1}\right )^{n+1} \ge p^{-1}a_{n+1} $$ For $x=p^{-1}a_{n+1}$ the inequality in question is $$\left (1+{x-1\over n+1} \right )^{n+1}\ge x$$ The last inequality follows from the Bernoulli inequality $$(1+y)^k\ge 1+ky,\quad y\ge -1$$ with $k=n+1$ and $y={x-1\over n+1}.$