Let $\langle x,y \rangle = x_1y_1 + 3x_2y_2 + 4x_3y_3 + x_1y_2 + x_2y_1 + x_1y_3 + x_3y_1 + x_2 y_3+x_3y_2$, prove pos. definiteness Let $$\langle x,y \rangle = x_1y_1 + 3x_2y_2 + 4x_3y_3 + x_1y_2 + x_2y_1 + x_1y_3 + x_3y_1 + x_2 y_3+x_3y_2$$, prove that $\langle x,y \rangle$ is positive definite.
I have simplified this to the following term: 
$$(x_1+x_2)^2+ (x_2 + x_3)^2 + x_2^2 + x_3^2 +2x_1x_2$$
How can I show that this inequality is greater than 0 for all $x \neq 0$?

We have to look whether the quadratic form
$$q(x):=\langle x,x\rangle=x_1^2+3x_2^2+4x_3^2+2(x_1x_2+x_2x_3+x_3x_1)$$
 is positive definite. To this end we complete the squares and write $q$ in the form
$$q(x)=(x_1+x_2+x_3)^2+2x_2^2+3x_3^2\ .$$
It is obvious that $q(x)\geq0$ for all $x$. Furthermore  $q(x)=0$  enforces $x_2=x_3=0$ and finally $x_1=0$. It follows that $q$ is indeed positive definite.
