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 $y=x-1$.

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    $\begingroup$ Well, it's essentially same as Pólya's, everything's just in terms of logarithm instead of exponential function, but the same idea and lemma is used. And you fixed the geometric mean instead of arithmetic to equal 1, which doesn't really affect anything. $\endgroup$ – gtrrebel Sep 7 '14 at 13:32
  • $\begingroup$ @gtrrebel - "..., which does not really affect anything". Well, it does affect a thing or two: the proof above is shorter and more straightforward. All variants of the proof according to Pólya (that I know of) perform some acrobatics. $\endgroup$ – chizhek Sep 8 '14 at 9:01
  • $\begingroup$ Compare it to the wikipedia version (en.wikipedia.org/wiki/…). Positivity of $x-1-\ln(x)$ corresponds to positivity of $e^{x-1}-x$, and your use of $x$ (that is, assuming geometric mean equals 1) corresponds to the division by $\alpha$ in wikipedia, a bit different tricks, but there's common idea behind them. And sure, wikipedia proof isn't for generalised case, but essentially nothing would change if the coefficients were added. It's a bit different, but I would definitely still call it a variant. $\endgroup$ – gtrrebel Sep 8 '14 at 10:44
  • $\begingroup$ @gtrrebel - Ok, I agree, it's a variant, within a broad definition of "variant". $\endgroup$ – chizhek Sep 8 '14 at 10:48
  • $\begingroup$ And yeah, in a way also Jensen and Pólya's ways are quite the same, it's about concavity of logarithm or convexity of exponential function, but here the connection is much clearer, I think. It's matter of opinion of course. $\endgroup$ – gtrrebel Sep 8 '14 at 10:53

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