As Gerry Myerson's question comment indicates, you have
$$\begin{equation}\begin{aligned}
& 2x_1^2 - 2x_1x_2 + 2x_2^2 - 2x_2x_3 + 2x_3^2 \\
& = x_1^2 + (x_1^2 - 2x_1x_2 + x_2^2) + (x_2^2 - 2x_2x_3 + x_3^2) + x_3^2 \\
& = x_1^2 + (x_1 - x_2)^2 + (x_2 - x_3)^2 + x_3^2 \\
& \ge 0
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Note it is $0$ only if $x_1 = x_2 = x_3 = 0$, so the result is always positive for any non-zero vector.
Regarding the stated result in your textbook, is it missing the $-$ signs or did you not put them in?
For $ k = 0, 1, \ldots, n-1$, consider the (horizontal) vector $v_k$ with $i$th coordinate $$ \sum \prod_{j=1, a_j \neq i}^{k} {x_{a_j}}.$$
For example, with $n = 3$, we have
$v_0 = (1, 1, 1)$,
$v_1 = (x_2 + x_3, x_3 + x_1, x_1 + x_2)$,
$v_2 = ( x_2x_3, x_3x_1, x_1x_2)$.
With $n = 4$, we have
$v_0 = (1, 1, 1, 1)$,
$v_1 = (x_2 + x_3 + x_4, x_3 + x_4 + x_1, x_4 + x_1 + x_2, x_1 + x_2 + x_3)$,
$v_2 = ( x_2x_3+x_3x_4+x_4x_2, x_3x_4+x_4x_1+x_1x_3, x_4x_1+x_1x_2+x_2x_4, x_1x_2 + x_2x_3 + x_3x_2)$,
$v_3 = (x_2x_3x_4, x_3x_4x_1, x_4x_1x_2, x_1x_2x_3)$.
Claim: $v_kA = (n-1-k) v_k$.
Proof: Expand it. A lot of the cross terms cancel out.
For example, with $v_0$, the column sum is $n-1$, so $v_0 A = (n-1) v_0$.
For example, with $v_{k-1}$, the numerators are all $\prod x_i$, and by looking at the denomninators, they cancel out to 0, so $v_{k-1} A = 0 $.
Do you see how we get $v_k A = (n-1-k)v_k$?
Corollary: The eigenvalues are $0, 1, 2, \ldots, n-1 $.
Best Answer
Note that we can show that a matrix is positive definite by looking at its $n$ upper left determinants. Note that for the matrix
$$\begin{bmatrix} 4 & 1 & 1\\1 & 2 & -1 \\ 1 & -1 & 3 \end{bmatrix}$$
We have $$\begin{vmatrix} 4 \end{vmatrix} = 4$$
and $$\begin{vmatrix} 4 & 1\\ 1 & 2 \end{vmatrix} = 7$$
and lastly
$$\begin{vmatrix} 4 & 1 & 1\\ 1 & 2 & -1\\ 1 & -1 & 3 \end{vmatrix} = 13$$
Since $4,7,13 > 0$ the matrix is positive definite.