Showing a function is not an inner product

Question

From Axler's Linear Algebra Done Right, 6.A 1,
Show that the function that takes $((x_1,x_2),(y_1,y_2)) \in \mathbb{R}^2 \times\mathbb{R}^2$ to $|x_1 y_1|+|x_2 y_2|$ is not an inner product on $\mathbb{R}^2$.

My attempt:
I will verify all the properties of inner products:
Positivity
$|x_1 x_1|+|x_2 x_2|$ is always $\geq0$ so this holds.

Definiteness
$\langle (x_1,x_2),(x_1,x_2)\rangle=|x_1 x_1|+|x_2 x_2|=0$ only if $|x_1 x_1|=|x_2 x_2|=0$ so $(x_1,x_2)=0$

Additivity in first slot
$\langle (x_1,x_2)+(z_1,z_2),(x_1,x_2)\rangle =\langle (x_1+z_1,x_2+z_2),(x_1,x_2)\rangle=|(x_1+z_1) y_1|+|(x_2+z_2)y_2|=|x_1 y_1|+|z_1 y_1|+|x_2 y_2|+|z_2 y_2|=\langle (x_1,x_2),(y_1,y_2)\rangle +\langle (z_1,z_2),(y_1,y_2)\rangle$

Homogeneity in first slot
$\langle \lambda (x_1,x_2),(y_1,y_2)\rangle=\langle (\lambda x_1,\lambda x_2),(y_1,y_2)\rangle =|\lambda x_1 y_1|+|\lambda x_2 y_2|=\lambda|x_1 y_1|+\lambda|x_2 y_2|=\lambda(|x_1 y_1|+|x_2 y_2|)=\lambda\langle (x_1,x_2),(y_1,y_2)\rangle$

Conjugate symmetry
$\overline {\langle (y_1,y_2),(x_1,x_2)\rangle}=\overline {|y_1 x_1|+|y_2 x_2|}=\overline{|y_1 x_1|}+\overline{|y_2 x_2|}=|y_1 x_1|+|y_2 x_2|=|x_1 y_1|+|x_2 y_2|$

It seems to me that everything holds so I am not sure why this is not an inner product.

Answer 1

HINT. $|a+b|\neq |a|+|b|$, $|\lambda a|\neq \lambda |a|$. Consider negative numbers.