I have already posted (most of) the present question as a (misplaced) answer to a question about understanding a particular proof of the AM-GM inequality. I sincerely hope I am not breaking the code of conduct on Stack Exchange.
Let $n\geq 1$ be an integer, let $a_1,\ldots,a_n>0$, and let $\lambda_1,\ldots,\lambda_n> 0$ satisfy $\lambda_1+\cdots+\lambda_n=1$. Then $a_1^{\lambda_1}\cdots a_n^{\lambda_n}\leq\lambda_1a_1+\cdots\lambda_na_n$, where the equality holds if and only if $a_1=\cdots=a_n$.
I will give my favorite proof of the generalized AM-GM inequality.
The embarrassing thing about this particular proof is that I cannot remember
whether I have seen it somewhere
or I hacked it up myself when I was fooling around thinking up different ways
(some of them extremely weird) of proving the inequality.
In case you have come across this proof, or its close relative, somewhere, anywhere,
please let me know.
(Yes, I know the proof by Pólya, I know the Jensen's inequality,
I know that the AM-GM inequality
is a manifestation of the concavity of the $\log$ function.)
All we need for the proof of the generalized AM-GM inequality is the following inequality:$ \newcommand{\RR}{\mathbb{R}}$
For every $x\in\RR$, $x>0$, we have $\,x-1\geq\ln x\,$, where the equality holds iff $\,x=1$.
The proof is simple: setting $f(x):=x-1-\ln x$ we have $f(1)=0$, and $f'(x)=1-x^{-1}$ is (strictly) negative for $0<x<1$ and is (strictly) positive for $x>1$.
My favorite proof of the generalized AM-GM inequality. $~$For every $x>0$ we have $$ \begin{aligned} (xa_1)^{\lambda_1}\cdots(xa_n)^{\lambda_n} ~&\:=\: x\cdot a_1^{\lambda_1}\cdots a_n^{\lambda_n}~, \\[1ex] \lambda_1(xa_1)+\cdots+\lambda_n(xa_n) ~&\:=\: x\cdot(\lambda_1a_1+\cdots\lambda_na_n)~. \end{aligned} $$ This means that it suffices to prove the inequality with $xa_1$, $\ldots$, $xa_n$ in place of $a_1$, $\ldots$, $a_n$ for any $x>0$ we choose. We choose $x=(a_1^{\lambda_1}\cdots a_n^{\lambda_n})^{-1}\!$, that is, we can assume that $a_1^{\lambda_1}\cdots a_n^{\lambda_n}=1$ and have to prove that then $1\leq \lambda_1a_1+\cdots+\lambda_na_n$. But this is easy: $$ \begin{aligned} \lambda_1a_1+\cdots+\lambda_na_n-1 ~&\:=\: \lambda_1(a_1-1) + \cdots +\lambda_n(a_n-1) \\[.5ex] ~&\:\geq\: \lambda_1\ln a_1 + \cdots + \lambda_n\ln a_n \\[.5ex] ~&\:=\: 0~, \end{aligned} $$ where the equality holds iff $a_1=\cdots=a_n=1$.$~$ Done.
Remark about my debate with gtrrebel in the comments to this question.
Let me clarify the question: Has anybody seen the particular proof given above,
or some proof that is very very close to it?
Note the "bim bam bom" scheme of the proof:
bim, we may prove any of rescaled inequalities instead of the given inequality;
bam, we choose the rescaled inequality that is simple in some sense;
bom, we give what amounts to a one-line proof of the simple inequality,
using the $\log$ inequality.
A proof that goes "bim bam boom (sounds of breaking glass) $\scriptsize\mathrm{bam}$"
and, moreover, does not use the $\log$ inequality, is no good as an answer.
We have here a kind of mathematical whodunit,
except that the final few pages are missing
so we cannot peek at them to find that it was the butler who did it.
Another remark about concavity of the $\log$ function. Because $\ln x$ is a strictly concave function of $x>0$, the diagram of $y=\ln x$ lies strictly below any of its tangents except, of course, at the point of contact. It is this property (more precisely, the analogous property for a general concave function that may not have tangents at all points) that is used in the proof of Jensen's inequality. In the proof above we got away using the single tangent $y=x-1$ of $y=\ln x$ at the point $(1,0)$. How come? Substitute $x/x_0$ for $x$ into the inequality $x-1\geq\ln x$ and you get $x/x_0-1+\ln x_0\geq\ln x$, which is the inequality corresponding to the tangent of $y=\ln x$ at the point $(x_0,\ln x_0)$; thus with the function $\ln$ we have all its tangents 'hiding' in the tangent $y=x-1$.
