A Question about a proof of the AM-GM inequality

Question

I'm a high school student, and I'm stuck on part 2 of this question. I would like some hints (not a full solution) on how to approach it:

1

The specific part of the question that confuses me is the significance of [Hint: The case $n=4$ is a good place to start.]. What about case $n=4$ makes it better over case $n=3$? Because I'm failing to see that.

What I've tried so far with case $n=4$ is that I've tried to show $((a+b+c+d)/4)^4 \geq abcd$ by expanding $(a+b+c+d)/4$. I don't think this is the right method though, for two reasons:

1. It doesn't seem to use the fact that $n=4$: I could have done $n=3$, with $((a+b+c)/3)^3 \geq abc$.

2. I don't see how I can generalize this method to an AM-GM inequality with any $n$ values.

So could someone give me a hint about the hint :)? I would really appreciate it if someone could explain how I could use this hint to solve this problem. Also, could you give me a hint only, and not a full solution? I still want solving the remaining problem to be a challenge.

Thanks in advance!

Answer 1

As others have noted, $n=4$ is a natural starting point because of Cauchy's double-then-climb-down inductive strategy. The $n=3$ case is reasonably easy on its own, viz.$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx),$$where the quadratic factor is $\ge0$ by Cauchy-Schwarz, but e.g. $n=7$ isn't as easy as $n=8$. But I disagree with

There are many proofs, none of them are completely straightforward.

In particular, we can use a "normal" proof by induction that doesn't snake through the values of $n$ in a nonstandard way. Perhaps the simplest proof (if the intended audience doesn't know calculus), if we restate the problem as proving $\prod_ia_i=1\implies\sum_ia_i\ge n$ for positive $a_i$, is to arrange $n+1$ terms in the inductive step so $a_1\ge1\ge a_2$, whence$$\begin{align}(a_1-1)(1-a_2)&\ge0\\\implies a_1+a_2&\ge 1+a_1a_2\\\implies\sum_ia_i&\ge 1+\underbrace{a_1a_2+\sum_{i\ge3}a_i}_{\ge n\text{ by inductive hypothesis}}\\&\ge n+1.\end{align}$$Edit: come to think of it, this approach is even simpler, albeit with induction on the number of steps it takes to make all values equal, rather than on $n$.

Answer 2

The simple part is $n=2$, as it follows from $(\sqrt a-\sqrt b)^2\ge 0$.

Show how the case for $n=2^{k+1}$ follows from the case $2^k$. By induction, conclude that the claim holds for all powers of $2$.

Now use "downward induction" to reach arbitrary $n$ from the next highest power of $2$.

Answer 3

Note that for any positives $a,$ $b$ and $c$ we have: $$\frac{a+b+c+\frac{a+b+c}{3}}{4}\geq\sqrt[4]{abc\cdot\frac{a+b+c}{3}}$$ Can you end it now?