Positive definiteness of a matrix Let ${x_1,x_2,...x_n}$ be positive numbers. Consider the matrix $C$ whose $(i,j)$-th entry is 
$$\min\left\{\frac{x_i}{x_j},\frac{x_j}{x_i}\right\}$$
Show that $C$ is non-negative definite (or positive semidefinite, meaning $z^t C z\geq 0$ for all $z\in \mathbb{R}^n$).When is $C$ positive definite?

We can suppose that $x_i\leq x_j$ for $i<j$ (otherwise just permute the components). As I'll show below, the determinant of your matrix is
$$\det C_n=\frac{x_2^2-x_1^2}{x_2^2}\frac{x_3^2-x_2^2}{x_3^2}\dots\frac{x_n^2-x_{n-1}^2}{x_n^2},\qquad(*)$$
which is $\geq 0$, and it is $>0$ iff all $x_i$'s are different. Notice that if you take the top left corner of $C$ with $k$ rows and columns, you get $C_k$, and $\det C_k\geq0$. Your matrix is therefore positive semidefinite (by Sylvester criterion), and it is positive definite iff all $x_i$'s are different.
Now we need to prove $(*)$. Let $D_n$ be $C_n$ with $i,k$-th element multiplied by $x_i x_j$, so that $\det D_n=\det C_n\times\prod_i x_i^2$. We want to show
$$\det D_n=x_1^2(x_2^2-x_1^2)\dots(x_n^2-x_{n-1}^2).$$
$D_n$ looks like
$$
\begin{pmatrix}
x_1^2 & x_1^2& x_1^2&x_1^2\\
x_1^2& x_2^2& x_2^2&x_2^2\\
x_1^2& x_2^2&x_3^2&x_3^2\\
x_1^2& x_2^2&x_3^2&x_4^2
\end{pmatrix}
$$
(for $n=4$ - I hope the pattern is clear), i.e.
$$
\begin{pmatrix}
a & a& a&a\\
a& b& b&b\\
a& b&c&c\\
a& b&c&d
\end{pmatrix}.
$$
($a=x_1^2,\dots,d=x_4^2$).
If we now subtract the first row from the others and then the first column from the others, we get
$$
\begin{pmatrix}
a & 0& 0&0\\
0& b-a& b-a&b-a\\
0& b-a&c-a&c-a\\
0& b-a&c-a&d-a
\end{pmatrix}.
$$
i.e. 
$$
\det
\begin{pmatrix}
a & a& a&a\\
a& b& b&b\\
a& b&c&c\\
a& b&c&d
\end{pmatrix}=
a\det \begin{pmatrix}
b-a& b-a&b-a\\
b-a&c-a&c-a\\
b-a&c-a&d-a
\end{pmatrix}
$$
Repeating this identity, we get
$$\det D_n=
\det
\begin{pmatrix}
a & a& a&a\\
a& b& b&b\\
a& b&c&c\\
a& b&c&d
\end{pmatrix}=
a\det \begin{pmatrix}
b-a& b-a&b-a\\
b-a&c-a&c-a\\
b-a&c-a&d-a
\end{pmatrix}=
a(b-a)\det
\begin{pmatrix}
c-b&c-b\\
c-b&d-b
\end{pmatrix}=
a(b-a)(c-b)(d-c)
$$
as we wanted to show.
There must be a more intelligent solution, but for the moment this one should do.
