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 $x = \left(\begin{array}{c}
x\\
y\\
z\ \end{array}\right)$
 A: 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.
A: One can see that the matrix $A$ has three positive eigenvalues. This means it is positive definite, and hence $x^{T} A x$ is positive for any non-zero $x$. 
A: Since $(x-y)^2 \geq 0$, it follows easily that $xy \leq \frac{1}{2}(x^2+y^2)$.  Similarly, $yz \leq \frac{1}{2}(y^2+z^2)$.  That gives you 
$$x^2+y^2+z^2-y(x+z) \geq \frac{1}{2}(x^2+z^2).  $$
The right-hand-side above is positive unless $x=z=0$.  But if $x=z=0$, then 
$$ x^2+y^2+z^2-y(x+z)=y^2,$$
and that will be positive unless $x=y=z=0$.
A: In my opinion it is easier to look at the Gershgorin circle theorem
to realize that this matrix $A$ has non negative eigenvalues (you do not need to compute them actually), in addition, since all the diagonal elements are strictly dominant excepting only one, then $A$ is invertible, therefore 0 is not an eigenvalue of $A$, hence $A$ is positive definite, which also means that $x^TAx > 0, x\neq0$
The nice part of this method is that it only takes you three sums xD
