Exercise 6.A.17 in "Linear Algebra Done Right 3rd Edition" by Sheldon Axler. I am worried if my solution is ok. I am reading "Linear Algebra Done Right 3rd Edition" by Sheldon Axler.

6.A.17 Prove or disprove: there is an inner product on $\mathbb{R}^2$ such that the associated norm is given by $$||(x,y)||=\max\{x,y\}$$ for all $(x,y)\in\mathbb{R}^2$.

I solved this exercise but I am worried if my solution is ok because this exercise appears to be unnaturally too easy.
My solution is the following:

If there is an inner product on $\mathbb{R}^2$ such that the associated norm is given by $$||(x,y)||=\max\{x,y\}$$ for all $(x,y)\in\mathbb{R}^2$, then $0<||(-1,-1)||=\max\{-1,-1\}=-1$.
This is a contradiction.

 A: As covered in the comments, you need absolute values to get what is called the $\infty$-norm: $$\|(x, y)\|_{\infty} = \max\{|x|, |y|\}.$$ This is the limit as $p \to \infty$ of the $p$-norm $$\|(x, y)\|_p = (|x|^p + |y|^p)^{1/p}$$
which generalizes the Euclidean norm ($p = 2$).
The rule for when a norm comes from an inner product is called the Parallelogram law which says that if $\| \cdot \|$ comes from an inner product, then for all $u, v$
$$\|u + v\|^2 + \|u - v\|^2 = 2(\|u\|^2 + \|v\|^2).$$
Moreover, if this identity holds for all $u$ and $v$, the inner product is recovered "by polarization" as
$$\langle u, v \rangle = \frac{\|u + v\|^2 - \|u - v\|^2}{4}.$$
So continuing with Axler's exercise, show that the $\infty$-norm (with the absolute values) does not satisfy the Parallelogram law. Slightly more challenging: show that the Parallelogram law holds for all $u, v$ for the $p$-norm if and only $p = 2$. I recommend thinking about an actual parallelogram.
A: Thank you very much Theo Bendit.

Prove or disprove: there is an inner product on $\mathbb{R}^2$ such that the associated norm is given by $$||(x,y)||=\max\{|x|,|y|\}$$ for all $(x,y)\in\mathbb{R}^2$.

My solution:

Assume that $\langle(x,y),(x,y)\rangle=\max\{|x|,|y|\}^2$.
Then, $\langle(1,0),(1,0)\rangle=\langle(0,1),(0,1)\rangle=1$.
Then, $1=\langle(1,1),(1,1)\rangle=\langle (1,0)+(0,1),(1,0)+(0,1)\rangle=\langle(1,0),(1,0)\rangle+2\langle(1,0),(0,1)\rangle +\langle(0,1),(0,1)\rangle=1+2\langle(1,0),(0,1)\rangle +1=2+2\langle(1,0),(0,1)\rangle$.
So, $\langle(1,0),(0,1)\rangle=-\frac{1}{2}$.
Then, $2=\langle(1,2),(1,2)\rangle=\langle (1,0)+(0,2),(1,0)+(0,2)\rangle=\langle(1,0),(1,0)\rangle+4\langle(1,0),(0,1)\rangle +4\langle(0,1),(0,1)\rangle=1-2+4=3$.
This is a contradiction.

