Stuck on proving two metrics are equivalent iff all the other related metrics are equivalent. Let $ d(x,y)= d_1 (x,y) + d_2 (x,y) $ and $ \delta (x,y) = \max(d_1 (x,y), d_2 (x,y)) $ and let $ d_1 $ and $ d_2 $ be metrics on $ M $.
Show that $ d $ and $ \delta $ are finer than $d_1$ and $d_2$.
Also, show that $d_1 \sim d_2$ iff $ d, d_1 , d_2 , \delta $ are equivalent between themselves.
I showed that both $d$ and $\delta$ are metrics on $M$ and that $d$ and $\delta$ are finer than $d_1$ and $d_2$.
Not sure how to prove the iff part.
 A: Well, to show that $d$ and $\delta$ are finer than $d_1, d_2$ it is sufficient to notice $d_1, d_2 \le d, \delta$ since then we have:
$$B_d(x, r), B_\delta(x, r) \subseteq B_{d_1}(x, r), B_{d_2}(x, r)$$
To see this, for example take $y \in B_d(x, r)$. Then we have $d_1(x, y) \le d(x, y) < r$, thus $y \in B_{d_1}(x, r)$. The other inclusions are similar.
Now assume $d_1 \sim d_2$.
Take an arbitrary ball $B_{d_1}(x, r)$.
Since $d_1\sim d_2$, there exists $r' > 0$ such that $B_{d_1}(x, r) \subseteq B_{d_2}(x, r')$.
So, if $d_1(x, y) < r$, we have $d_2(x, y) < r'$, so $d(x, y) < r + r'$ and $\delta(x, y) < \max\{r,r'\}$.
Thus, $B_{d_1}(x, r) \subseteq B_{d}(x, r + r')$ and $B_{d_1}(x, r) \subseteq B_\delta(x, \max\{r,r'\})$.
Similarly, take the ball $B_{d_2}(x, r)$. Since $d_1\sim d_2$, there exists $r'' > 0$ such that $B_{d_2}(x, r) \subseteq B_{d_1}(x, r'')$.
So, if $d_1(x, y) < r$, we have $d_2(x, y) < r''$, so $d(x, y) < r + r''$ and $\delta(x, y) < \max\{r,r''\}$.
Thus, $B_{d_2}(x, r) \subseteq B_{d}(x, r + r'')$ and $B_{d_1}(x, r) \subseteq B_\delta(x, \max\{r,r''\})$.
Thus, the open balls of $d$ and $\delta$ nest with the open balls of $d_1$ and $d_2$ so all the metrics in question are equivalent.
