The $A_m≥G_m$ has a whole lot of proofs, the large majority I've seen by induction. Currently the quickest one that I've seen is the following: (note $A_n=\frac{a_1+a_2+a_3+...+a_n}{n}, G_n=\sqrt[n]{a_1a_2a_3...a_n}$ where $a_k$ is a positive real number).
Consider the geometric series: $1+(1+x)+(1+x)^2+(1+x)^3+...+(1+x)^{n-1}=\frac{(1+x)^n-1}{x}$ for all integers $n≥0$.
If $x≥0$ then clearly $1+(1+x)+(1+x)^2+(1+x)^3+...+(1+x)^{n-1}≥n$ which we then substitute into the expression to get $(1+x)^n≥1+nx$, if $-1≤x<0$ then $1+(1+x)+(1+x)^2+(1+x)^3+...+(1+x)^{n-1}≤n$ which then substitutes re-arranges to $(1+x)^n≥1+nx$ again, noting that the sign flips when the $x$ in the denominator is multiplied across because it is assumed to be negative.
Hence we have shown $(1+x)^n≥1+nx$ is true for $x≥-1$. Then for $n=n+1$ substitute $x=\frac{A_{n+1}}{A_{n}}-1$ to derive the result $\left(\frac{A_{n+1}}{A_{n}}\right)^{n+1}≥\frac{a_{n+1}}{A_n}$.
The proof then finishes with induction with the assumption being $A_k≥G_k$ the the $k+1$ step being:
$LHS=G_{k+1}$
$=\sqrt[k+1]{a_1a_2a_3...a_{k+1}}$
$=(\sqrt[k]{a_1a_2a_3...a_{k}})^{\frac{k}{k+1}}\sqrt[k+1]{a_{k+1}}$
$≤(A_k)^{\frac{k}{k+1}}\sqrt[k+1]{a_{k+1}}$ by assumption
$≤\left(\sqrt[k]{\frac{A_{k+1}^{k+1}}{a_{k+1}}}\right)^{\frac{k}{k+1}}\sqrt[k+1]{a_{k+1}}$ using the proved inequality rearranged for $A_k$
$=A_{k+1}$ which completes the proof.
I was wondering what is the absolute quickest way (the least amount of steps that can still be followed without any extra work between those steps) that I can prove $A_n≥G_n$ from the ground up (maybe without induction)? That is, not using any other inequality results without proof, (this proof used Bernoulli's inequality but that was relatively quick to prove) and not using any advanced techniques (to me) such as matrices or something.
$\ln$ is concave then : $$\ln \dfrac{a_1 + \cdots + a_n}{n} \geq \dfrac{\ln a_1 + \cdots + \ln a_n}{n} = \ln \sqrt[n]{a_1 \cdots a_n}$$ And we are done.