I'm trying to prove the following: $\sqrt[4]{abcd} \le \frac{a+b+c+d}{4}$.
While I managed to prove base case $n=1$ and $n=2$;
$(\sqrt{a}-\sqrt{b})^2 \geqslant 0$
$a-2\sqrt{ab}+b \geqslant 0$
$a+b \geqslant 2\sqrt{ab}$
$\sqrt{ab} \le \frac{a+b}{2}$
I am unsure how to proceed for $n=4$. Is there a way to continue proving 4 variables using induction, or is it better to use Cauchy-Schwarz Inequality to go about the proof?
You can prove AM-GM using Jensen's inequality which states that if $f$ is continuous and convex (or equivalently, concave up), then we know for $x_1, x_2, ..., x_n$ and $a_1, a_2, ..., a_n$ that $f(\frac{a_1x_1 + ... + a_nx_n}{a_1 + ... + a_n}) \leq \frac{a_1 f(x_1) + ... + a_n f(x_n)}{a_1 + ... + a_n}$. This may seem complicated, but we can think of the $a_i$ as weights being applied before or after $f$. Since $f$ is convex, the point $\frac{a_1x_1 + ... + a_nx_n}{a_1 + ... + a_n}$ will be somewhere in the middle of the $x_i$. Then, when $f$ is applied to this value, we get something less than a value which is somewhere in the middle of the $f(x_i)$. It can be shown that Cauchy-Schwarz is actually a specific case of Jensen's inequality.
We actually apply this as follows: Consider $x_1, x_2, ..., x_n \in \mathbb{R}$. Let $y_i = \text{ln}(x_i)$, let $a_i = \frac{1}{n}$, and let $f(x) = e^x$, which is convex/concave up. Thus, by Jenson's inequality, $e^{\frac{1}{n}(y_1 + ... + y_n)} \leq \frac{1}{n}(e^{y_1} + ... + e^{y_n})$. By the construction of $y_i$, we have that $e^{y_i} = x_i$. Additionally, $y_1 + ... + y_n = \text{ln}(x_1) + ... + \text{ln}(x_n) = \text{ln}(x_1 x_2 ... x_n)$. So plugging everything in and simplifying, we get $e^{\frac{1}{n}(y_1 + ... + y_n)} = (x_1 ... x_n)^\frac{1}{n} \leq \frac{1}{n}(x_1 + ... + x_n)$. At this point, we can let $n = 4$, to get that $n = 4$ to get your result.
Alternative approach.
Lemma-1
Given:
To Prove: $\sum_{i=1}^n x_i > n.$
Proof by Induction
$n = 2:$
Without Loss of Generality (WLOG), $x_1 = 1 - k, ~0 < k < 1.$
Then $\displaystyle x_2 = \frac{1}{x_1} = \frac{1}{1 - k} = 1 + k + \frac{k^2}{1 - k} \implies $ $\displaystyle x_1 + x_2 = 2 + \frac{k^2}{1 - k} > 2.$
$n = (N+1):$
Assume conjecture true when $n = N$.
Given $x_1, \cdots, x_N, x_{(N+1)} \in \mathbb{R+},~$
such that they are not all equal and $\prod_{i=1}^{(N+1)} x_i = 1.$
WLOG $~x_1 < 1 < x_{(N+1)}.$
Let $b = x_1 \times x_{(N+1)}.$
By induction hypothesis,
since $\displaystyle b \times \prod_{i=2}^N x_n = 1~$,
$~~b + \sum_{i=2}^N x_i > N.$
Edit
Minor flaw in the above analysis, which (fortunately) doesn't invalidate the analysis at the end of the proof to Lemma 1.
As $b, x_2, x_3, \cdots, x_N$ are defined, it is possible that
these $N$ elements are all equal to $1$.
Therefore, the correct intermediate conclusion is that
$b + \sum_{i=2}^N x_i \geq N.$
Since $~x_1 < 1 < x_{(N+1)},~~$ $0 > [x_1 - 1]~[x_{(N+1)} - 1] = -x_1 - x_{(N+1)} + 1 + b \implies$
$x_1 + x_{(N+1)} > 1 + b \implies \sum_{i=1}^{N+1} x_i > (N+1).$
Notes:
When proving that the geometric mean is greater than or equal to the arithmetic mean, you can assume that each of the numbers is non-negative. Otherwise, the conjecture is false. For example: the geometric mean of $(-4)$ and $(-9)$ will be construed to be either $\pm 6$, both of which are greater than $\frac{(-4) + (-9)}{2}.$
Consequently, you can also assume that each of the numbers is non-zero, because otherwise the geometric mean would be $0$. Therefore, by presumption, each of the numbers is positive.
The situation is trivial when all of the numbers are equal, since then the geometric mean and arithmetic mean are then clearly equal. Therefore, WLOG, the numbers are all positive, and not all equal.
Given $\displaystyle x_1,x_2, \cdots, x_n, ~\text{let}~ G = \left(\prod_{i=1}^n x_i\right)^{(1/n)}.$
For $i \in \{1,2,\cdots,n\},~$ let $\displaystyle a_i = \frac{x_i}{G}.$
Then, $\displaystyle\prod_{i=1}^n a_i = 1.$
Therefore, by Lemma-1, $\sum_{i=1}^n a_i > n.$
Thus $\displaystyle \frac{x_1}{G} + \cdots + \frac{x_n}{G} > n \implies \frac{x_1 + \cdots + x_n}{n} > G.$
