If $M$ is a compact Riemannian manifold and $g$ and $\tilde{g}$ are metrics on $M$, then $\frac{1}{C} g \leq \tilde{g} \leq C g$ for $C > 1$

You can prove this in a more direct way. It looks like the proof that in a finite dimensional vector space, all norms are equivalent.

Let $S_gM$ be the unit sphere bundle of $(M,g)$, that is $S_gM = \{ (p,v)\in TM | g_p(v,v)=1 \}$. If $M$ is compact, then $S_gM$ is compact too. The smooth function $f$ on $TM$ defined by $f(p,v)= \tilde{g}_p(v,v)$ is then continuous restricted to $S_gM \subset TM$. Notice $f$ is positive, as every $v\in S_gM$ is non-zero. By compactness, there exist $m,M >0$ such that $m\leqslant f(p,v) \leqslant M$ on $S_gM$. You can chose some constant $C>1$ such that $\frac{1}{C} \leqslant m \leqslant M \leqslant C$, so that on $S_gM$, $\frac{1}{C} \leqslant \tilde{g}_p(v,v)\leqslant C$. By the very definition of $S_gM$, we have that for every $(p,v)\in S_gM$, $$\frac{1}{C}g_p(v,v)\leqslant \tilde{g}_p(v,v) \leqslant Cg_p(v,v)$$ Now, the homogeneity of quadratic forms show that this inequality is true on all $TM$.