Let $a_1,...,a_n>0$. The arithmetic mean is defined by $A(a_1,...,a_n) =\frac{a_1+...+a_n}{n}$ and the geometric mean by $G(a_1,...,a_n)=\sqrt[n]{a_1\cdot ...\cdot a_n}$.

Let $S(n)$ be the statement: $$\forall a_1,...,a_n >0: G(a_1,...,a_n) \leq A(a_1,...,a_n)$$

a) Prove $S(2)$ is true

b) Let $n\geq 2$. Prove that if $S(2)$ and $S(n)$ are true, then also $S(2n)$.

c) Prove for all $n\geq 2$ that if $S(n+1)$ is true, then $S(n)$ is true. [Hint: use $G(a_1,...,a_n,G(a_1,...,a_n))=G(a_1,...,a_n)$].

d) Explain why this shows that $S(n)$ is true for all $n\geq 2$

a) This is easy to prove from $(a_1 - a_2)^2\geq 0$

After this, I am completely stumped. Any hints or ideas in which to build on?


To $b):$

Assume $S(2)$ and $S(n)$ hold. Then,

$$\sqrt[2n]{a_1\cdots a_{2n}}=\sqrt{\sqrt[n]{a_1\cdots a_n} \sqrt[n]{a_{n+1}\cdots a_{2n}}}\underbrace{\le}_{S(2)} \frac{\sqrt[n]{a_1\cdots a_n }+\sqrt[n]{a_{n+1}\cdots a_{2n} }}{2} \\ \underbrace{\le}_{S(n)} \frac{\displaystyle \frac{a_1+\cdots +a_n }{n}+\frac{a_{n+1}+\cdots +a_{2n}}{n} }{2} =\frac{a_1+\cdots +a_{2n}}{2n},$$ which shows that $S(2n)$ holds.

To $c):$

Assume that $S(n+1)$ holds. Then, using the hint,

$$\sqrt[n]{a_1\cdots a_n}=\sqrt[n+1]{a_1\cdots a_n \sqrt[n]{a_1\cdots a_n}}\underbrace{\le}_{S(n+1)} \frac{a_1+\cdots a_n +\sqrt[n]{a_1\cdots a_n}}{n+1} \\ \implies (n+1)\sqrt[n]{a_1\cdots a_n}\le a_1+\cdots a_n +\sqrt[n]{a_1\cdots a_n}\\ \implies \sqrt[n]{a_1\cdots a_n}\le \frac{a_1+\cdots +a_n }{n},$$ which shows that $S(n)$ holds.

To $d):$

Since $S(2)$ holds it follows from $b)$ that $S(4)$ holds. Prove by induction that $S(2^n)$ holds for any $n.$ Then conclude from $c).$

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For b):

$$\begin{align} G(a_1, \cdots, a_{2n}) &= (\prod_{k=1}^{n}\sqrt{a_{2k-1}a_{2k}})^{1/n} \\ &\leq (\prod_{k=1}^{n}\dfrac{a_{2k-1} + a_{2k}}{2})^{1/n} \\ &\leq \dfrac{\sum_{k=1}^n \dfrac{a_{2k-1} + a_{2k}}{2}}{n} = A(a_1, \cdots, a_{2n}) \end{align}$$

c) can be proved by similar manipulation using the hint.

For d): by b), we have $S(n)$ is true for all $n = 2^k$, then apply c) to conclude for other $n$

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