Show that $\langle x,x\rangle > 0$ if $x \neq0$ Let $x=(x_1,x_2)$ and $ y=(y_1,y_2)$ be vectors in the vector space $C^2$ over $C$ and define $\langle \cdot,\cdot\rangle : C^2 \times C^2 \rightarrow C$ by
$ \langle x,y\rangle =3x_1 \overline{y_1}+(1+i)x_1 \overline{y_2}+(1-i)x_2 \overline{y_1}+x_2 \overline{y_2} $
Show that $\langle x,x\rangle > 0$ if $x \neq0$.
So what I have so far is:
$\begin{aligned} \langle x,x\rangle &= 3x_1 \overline{x_1}+(1+i)x_1 \overline{x_2}+(1-i)x_2 \overline{x_1}+x_2 \overline{x_2} \\
&= 3|x_1|^2 + |x_2|^2 +x_1 \overline x_2 + ix_1\overline x_2 - ix_2 \overline x_1 +x_2 \overline x_1 \\
&=3|x_1|^2 + |x_2|^2 + (1+i)(x_1 \overline x_2 - i\overline x_1 x_2)
\end{aligned}$
Now $3|x_1|^2 + |x_2|^2 >0$ if $x \neq0$, but I don't know how to show that $(1+i)(x_1 \overline x_2 - i\overline x_1 x_2) >0$ as well.
Any guidance would be appreciated.
Thank you.

\begin{align*}
\langle x,x\rangle &= 3|x_1|^2 + |x_2|^2 + 2 \operatorname{Re}((1+i)x_1 
\overline{x_2}) \\
&\ge 3|x_1|^2 + |x_2|^2 - 2 |(1+i)x_1 
\overline{x_2})| \\
&= 3|x_1|^2  - 2 \sqrt 2 |x_1| |x_2|+ |x_2|^2
\end{align*}
So,
$$\langle x,x\rangle = 0 \implies 3|x_1|^2  - 2 \sqrt 2 |x_1| |x_2|+ |x_2|^2=0$$
Now, calculate the determinant of
$$3|x_1|^2  - 2 \sqrt 2 |x_1| |x_2|+ |x_2|^2$$
to see that this can be $0$ if and only if $x_1=x_2=0$.
