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.

Let $x_1=a+bi$,  $x_2=c+di$.  Then
$$(1+i)(a+bi)(c-di)+(1-i)(a-bi)(c+di)=$$
$$(1+i)((ac+bd)+(bc-ad)i)+(1-i)((ac+bd)+(ad-bc)i)=$$
$$((ac+bd)-(bc-ad))+((ac+bd)+(bc-ad))i+((ac+bd)-(ad-bc)+((ad-bc)-(ac+bd))i=$$
$$(ac+bd-bc+ad+ac+bd-ad+bc)+(ac+bd+bc-ad+ad-bc-ac-bd)i=$$
$$2ac+2bd$$
So it isn't always greater than 0 by itself,  you need to show that
$$3(a^2+b^2)+(c^2+d^2)>-(2ac+2bd)$$
as long as at least one of $a,b,c,d\neq 0$
Can you finish?
