How do you prove the inequality $x^2+y^2+z^2-y(x+z)>0$? This actually comes out of a matrix multiplication of $x^TAx$, but when you multiply it out, you get the following.
$x^2+y^2+z^2-y(x+z)>0$
I just can't figure out how to actually prove that that inequality holds for all $x,y,z\neq0$.  Also, note that this is for a linear algebra class, so I don't think you're allowed to use derivatives.  Also, in case it is important, here are the original matrices used to get that inequality. 
$A=$ $
\left( \begin{array}{ccc}
2 & -1 & 0 \\
-1 & 2 & -1 \\
0 & -1 & 2 \end{array} \right)$
and $x = \left(\begin{array}{c}
x\\
y\\
z\ \end{array}\right)$

The inequality is
$$\frac12\left(x^2 + (x-y)^2 + (y-z)^2 + z^2 \right) > 0.$$
It is easy to see that that holds for all real $(x,y,z) \neq (0,0,0)$.
A more linear-algebra-y way is to consider the minors of $A$; we have $2 > 0$,
$$\begin{align}
\det \begin{pmatrix}2 & -1\\-1 & 2 \end{pmatrix} &= 2^2 - (-1)^2 = 3 > 0\\
\det \begin{pmatrix}2 & -1 & 0\\ -1 & 2 & -1\\ 0 & -1 & 2 \end{pmatrix} &= 8 + 0 + 0 - 0 - 2 - 2 = 4 > 0,
\end{align}$$
so $A$ is positive definite.
