The Geometric Mean as a Minimum. This is problem 8.5 from the book "Cauchy Schwarz Master Class":
Problem: Prove that the geometric mean has the representation $$\left(\prod_{k=1}^n a_k \right)^{1/n} = \min\left\{\frac{1}{n} \sum_{k=1}^n a_kx_k:(x_1,\cdots,x_n)\in D\right\},$$
where $a_k \geq 0$ and $D$ is the region of $\mathbb{R}^n$ defined by $$D = \left\{(x_1,\cdots,x_n): \prod_{k=1}^n x_k =1, x_k \geq 0, k = 1,2, \cdots, n\right\}.$$
Question: The author proceeded to use the AM-GM inequality to show that
$$\left(\prod_{k=1}^n a_k \right)^{1/n}  = \left(\prod_{k=1}^n a_kx_k \right)^{1/n} \leq \frac{1}{n}\sum_{k=1}^n a_kx_k,$$and showed that equality occurs when $a_kx_k$ does not depend on $k$. But then the author said that picking $x_k = \frac{a_k}{(\prod\limits_{k=1}^n a_k)^{1/n}}$ yields equality in the AM-GM bound. However, when I plugged this in, $a_kx_k$ is constant iff $a_k$ is constant, which is not given. Am I missing anything here?
 A: Looks like the  equality-yielding definition for $x_k$ is upside down, or an exponent $-1$ is missing on one side. The assertion $a_kx_k=c$ for all $k=1,\ldots n$ leads to
$$
\prod_{k=1}^n a_k\prod_{k=1}^n  x_k = c^n.
$$ Since $\prod_{k=1}^n  x_k=1$, this gives $$c=\left(\prod_{k=1}^n  a_k\right)^{1/n}$$ and
$$
x_k = \frac c{a_k}=\frac{(\prod\limits_{k=1}^n a_k)^{1/n}}{a_k}.$$
A: In the usual AM-GM inequality $$\left(\prod_{k=1}^n b_k \right)^{1/n} \le \frac{1}{n} \sum_{k=1}^n b_k$$ (where I have used $b_k$ to avoid confusion), equality occurs if and only if $b_k = b$ for some constant $b \ge 0$.  This is what the author means by "$a_k x_k$ does not depend on $k$."  But that does not imply that the sequence $\{a_k \}_{k=1}^n$ is itself constant.  It implies the sequence of the products $\{a_k x_k \}_{k=1}^n$ is constant, where the $x_k$ can be thought of as "weights" on the $a_k$.  The $a_k$ are fixed; the goal is to find a corresponding sequence of weights $x_k$ that results in equality.
To illustrate, let us consider a concrete example.  Let $n = 5$ and $a_k = k^2$ for $1 \le k \le n$.  Then the usual AM-GM statement is $$(1 \cdot 4 \cdot 9 \cdot 16 \cdot 25)^{1/5} \le \frac{1 + 4 + 9 + 16 + 25}{5},$$ which of course is true, and the inequality is strict.  Now can we find a sequence $x_k$ such that $$(1 \cdot 4 \cdot 9 \cdot 16 \cdot 25)^{1/5} = \frac{1x_1 + 4x_2 + 9x_3 + 16x_4 + 25x_5}{5}?$$  Certainly.  In fact, there are many possible solutions, but we want one in which the product $x_1 x_2 x_3 x_4 x_5 = 1$.  To do this, we observe that when this product of weights equals $1$, the LHS can be written $$(1 x_1 \cdot 4 x_2 \cdot 9 x_3 \cdot 16 x_4 \cdot 25 x_5)^{1/5}.$$  And now, by AM-GM applied to the sequence of prodcuts, equality is attained when $1x_1 = 4x_2 = 9x_3 = 16x_4 = 25x_5$; this gives us $$x_k = \frac{c}{a_k}$$ for some constant $c$.  Then since we require this to equal $1$, we have $$c^5 = a_1 a_2 a_3 a_4 a_5,$$ hence $$x_k = \frac{(a_1 a_2 a_3 a_4 a_5)^{1/5}}{a_k}.$$
Thus, the author has incorrectly taken the reciprocal of the required weight, and you can see the incorrect choice leads to an expression on the RHS of the form $\frac{1}{c^5} (a_1^2 + a_2^2 + \cdots + a_n^2),$ which clearly does not work.
