Show that $\langle ,\rangle$ defined for all $\mathbf x=[x_1,x_2]^\top\in\mathbb R^2$ and $\mathbf y=[y_1,y_2]^\top\in\mathbb R^2$ by $$\langle \mathbf x,\mathbf y\rangle := x_1y_1-(x_1y_2+x_2y_1)+2(x_2y_2)$$ is an inner product.
Solution: We check it by definition.
First, we check it is symmetric. Note that $$\langle \mathbf x,\mathbf y\rangle = x_1y_1-(x_1y_2+x_2y_1)+2(x_2y_2)$$ and $$\langle \mathbf y,\mathbf x\rangle = y_1x_1-(y_1x_2+y_2x_1)+2(y_2x_2),$$it is clear that $\langle \mathbf x,\mathbf y\rangle=\langle \mathbf y,\mathbf x\rangle$.
Second, we check it is bilinear. Let $\mathbf z=[z_1,z_2]^\top$, then for any real numbers $a,b$ we have\begin{align*}\langle a\mathbf x+b\mathbf y,\mathbf z\rangle=&\ \langle [ax_1+by_1,ax_2+by_2]^\top,[z_1,z_2]^\top\rangle \\ =&\ (ax_1+by_1)z_1-(ax_1+by_1)z_2-(ax_2+by_2)z_1+2(ax_2+by_2)z_2\\=&\ (ax_1z_1-ax_1z_2-ax_2z_1+2ax_2z_2)+(by_1z_1-by_1z_2-by_2z_1+2by_2z_2)\\=&\ a(x_1z_1-x_1z_2-x_2z_1+2x_2z_2)+b(y_1z_1-y_1z_2-y_2z_1+2y_2z_2)\\=&\ a\langle \mathbf x,\mathbf z\rangle+b\langle \mathbf y,\mathbf z\rangle.\end{align*} Use the symmetry shown above, we have\begin{align*}\langle \mathbf x,a\mathbf y+b\mathbf z\rangle=&\ \langle a\mathbf y+b\mathbf z,\mathbf x\rangle \\ = &\ a\langle \mathbf y,\mathbf x\rangle + b\langle \mathbf z,\mathbf x\rangle \\ = &\ a\langle \mathbf x,\mathbf y\rangle + b\langle \mathbf x,\mathbf z\rangle.\end{align*}In the second equality, we used the linearity shown above on the first component. The last equality comes from the symmetry of $\langle ,\rangle$.
Finally, we prove it is positive-definite. We have \begin{align*}\langle \mathbf x,\mathbf x\rangle=&\ x_1x_1-(x_1x_2+x_2x_1)+2x_2x_2\\ = &\ x_1^2-2x_1x_2+2x_2^2=(x_1-x_2)^2+x_2^2,\end{align*}which is always non-negative and can only be zero if $x_2=0$ and $x_1=x_2$. This confirms it is positive-definite.