functional analysis – Equivalent norms in the intersection

Let $V$ be a vector space. Two norms $|cdot |_1,|cdot |_2: V longrightarrow mathbb{R}$ in $V$ are said equivalent if there exists $a,b>0$ such that
$$
a|u|_1 leq |u|_2 leq b |u|_1,; forall ; u in V.
$$

Now,, consider $X=(X, |cdot|_X)$ and $Y=(Y, |cdot|_Y)$ two normed spaces. It is easy to show that $Z:= X cap Y$ is a normed space with norm $|cdot|_Z: Z longrightarrow mathbb{R}$ given by
$$
|u|_Z=|u|_X+|u|_Y,; forall ; u in Z.
$$

Question. If, for some some $alpha, beta>0$, we consider ${rm N}: Z longrightarrow mathbb{R}$ as
$$
{rm N}(u)=alpha |u|_X +beta |u|_Y,; forall ; u in Z
$$

then ${rm N}$ defines a equivalent norm (in $Z$) with respect the norm $|cdot|_Z$?

I don’t see, for instance, how to prove that
$$
alpha |u|_X +beta |u|_Y leq gamma( |u|_X + |u|_Y),; forall ; u in Z tag{1}
$$

for some $gamma>0$. Can I to use some property of $max$ or $min$? Or, can I only prove $(2)$ instead of the equivalence of norms? The equivalence is more stronger than $(2)$.