Inequality: $x^2+y^2+z^2+t(xy+yz+zx) \geq 0$ Prove that $x^2+y^2+z^2+t(xy+yz+zx) \geq 0$ for any $x,y,z \in \mathbb{R}$ and any $t \in [-1,2].$
One try:
for $t=-1$: $x^2+y^2+z^2-xy-yz-zx \geq 0$ is true .
for $t=2$: $x^2+y^2+z^2+2(xy+yz+zx) \geq 0$ is true
also for $t=0$.
But how can prove for $t \in (-1,0)$ and $t \in (0,2)$. 
Also I don't know if what I did is the right step. 
Thanks :) 

Note that
\begin{eqnarray}
x^2+y^2+z^2+t(xy+yz+zx)&=&\frac{2-t}{3}(x^2+y^2+z^2-xy-yz-zx)\\
&&+\frac{1+t}{3}(x^2+y^2+z^2+2xy+2yz+2yz) \, .
\end{eqnarray}
Your previous work shows that both terms on the right-hand side are always non-negative for $t \in [-1, 2]$; thus the left-hand side is as well.
