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.
 A: 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.
A: GEdgar gave a beautiful geometric answer, but I'd like to give one more example since OP asked for it in the comments.
The only property that fails is the triangle inequality.
Verifying the first two:

*

*Let $x\in V$. Then $\|x\|=0$ iff $\|x\|_1=0$ or $\|x\|_2=0$ which occurs iff $x=0$, since $\|\cdot\|_1,\|\cdot\|_2$ are norms.


*Let $x\in V$ and $\alpha\in\mathbb{R}$ a scalar. Then $\|\alpha x\|=\min\{\|\alpha x\|_1,\|\alpha x\|_2\}=\min\{|\alpha|\cdot\|x\|_1,|\alpha|\cdot\|x\|_2\}=|\alpha|\cdot\min\{\|x\|_1,\|x\|_2\}=|\alpha|\cdot\|x\|$.
Example where the triangle inequality fails:
Take $V=\mathbb{R}^2$ and consider the norms defined by $\|(x,y)\|_1=\frac56(|x|+|y|)$ and $\|(x,y)\|_2=\sqrt{x^2+y^2}$. Denote by $\|\cdot\|$ the minimum of the two norms.
Consider the points $p=(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ and $q=(0,\frac{6}{5})$ and let $\alpha=3/4$. Note (by doing the calculations) that: $\|p\|_1<\|p\|_2=1$ and that $\|q\|_2<\|q\|_1=1$. Now if the triangle inequality was true, we would have:
$$\|\alpha\cdot q+(1-\alpha)p\|\le\frac{3}{4}\|q\|+\frac{1}{4}\|p\|<\frac{3}{4}+\frac{1}{4}=1.$$
However,
$$\|\alpha q+(1-\alpha)p\|_1=\frac{5}{6}(\frac{\sqrt{2}}{8}+\frac{18}{20}+\frac{\sqrt{2}}{8})\approx1.04>1$$
and
$$\|\alpha q+(1-\alpha)p\|_2=\sqrt{\bigg(\frac{\sqrt{2}}{8}\bigg)^2+\bigg(\frac{18}{20}+\frac{\sqrt{2}}{8}\bigg)^2}\approx1.09>1$$
hence the minimum of these two numbers (which is approximately 1.04) is greater than 1, a contradiction.
