Let $\| x \|_1$ and $\| x \|_2$ be norms on a vector space $V$. Is $\|x\|=\min\{\| x \|_1,\| x \|_2\}$ necessarily a norm? Question: Let $\| x \|_1$ and $\| x \|_2$ be norms on a vector space $V$. Is $\|x\|=\min\{\| x \|_1,\| x \|_2\}$ necessarily a norm?
Attempt:
Since $\| x \|_1$ and $\| x \|_2$ are themselves norms, they satisfy the following properties: nonnegativity, positive definiteness, absolute homogeneity, and the triangle inequality. Thus, $\min\{\| x \|_1,\| x \|_2\} \geq 0$ since both $\| x \|_1 \geq 0$ and $\| x \|_2 \geq 0$, $\min\{\| x \|_1,\| x \|_2\} =0\Longleftrightarrow$ if at least one of $\| x \|_1$, $\| x \|_2$ are $0$, i.e. if $\min\{\| x \|_1,\| x \|_2\}=0$. Let $\alpha, \beta \in \mathbb{R}$. Then, $\min\{\| \alpha x \|_1,\| \beta x \|_2\}$ is either $\alpha \| x \|_1$ or $\beta \| x \|_2$. I'm not sure if this is correct.
Also, I am stuck on proving the triangle inequality part.

Example...
In $V = \mathbb R^2$, let
$$
\|(x,y)\|_1 = 4(x+y)^2+(x-y)^2,\qquad
\|(x,y)\|_2 = (x+y)^2+4(x-y)^2.
$$
Here is the unit ball for $\|\cdot\|_1$

Here is the unit ball for $\|\cdot\|_2$

Finally, here is the set where the minimum is equal to $1$:

Note that this is not a convex body; so this minimum is not a norm.
