<s>[INST] 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 $x = \left(\begin{array}{c}
x\\
y\\
z\ \end{array}\right)$
 [/INST] 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 $x=y=z=0$.
</s>