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 $C > 1$ I am reading Nonlinear Analysis on Manifolds: Sobolev Spaces and Inequalities by Emmanuel Hebey and he stated on page $22$:

Let $M$ be a compact manifold endowed with two Riemannian metrics $g$ and $\tilde{g}$. As one can easily check, there exists $C > 1$ such that
$$\frac{1}{C} g \leq \tilde{g} \leq C g$$
on $M$, where such inequalities have to be understood in the sense of the bilinear forms.

I would like to help to prove this, because I can not give a satisfactory proof with my attempt, but I put it below to show my effort. I also would like to apologize if my proof is very detailed, but I would like to see if I understood very well the argument and what hypothesis are used and how they are used.
It is sufficient to prove that $\frac{1}{C} \delta_j^i \leq \tilde{g}_{ij} \leq C \delta_j^i$ on $M$ for some constant $C > 1$. Suppose that $\tilde{g}$ is a Riemannian metric which is geodesic normal coordinates at $p$ without loss of generality because if the inequalities above are proved, then the inequalities are true for the metric $\tilde{g}$ which is not geodesic normal coordinates at $p$ only changing $C$ by $\frac{C}{A}$, where $A$ denotes the Jacobian of the change of the coordinates. Now, consider $M$ connected (the author assumes in the beginning of the book that manifolds are connected, I think this is used here to define the next metric on $M$) and endowed with the metric $d(p,q) := \inf \left\{ l(\alpha) \ ; \ \alpha \ \text{is a piecewise differentiable curve joining} \ p \ \text{to} \ q \right\}$. Recall that the Riemannian metric $\tilde{g}$ is smooth in the sense that the map
\begin{align*}
\tilde{g}: (M,d) &\longrightarrow (\mathscr{L}^2(T_pM \times T_pM, \mathbb{R}),||\cdot||_{op})\\
p &\longmapsto \tilde{g}(p)
\end{align*}
is smooth ($||\cdot||_{op}$ denotes the operator norm over $\mathscr{L}^2(T_pM \times T_pM, \mathbb{R})$), in particular, the map above is a continuous map defined over a compact metric space, then it is uniformly continuous. This part I am stuck, but I want to  define a norm $||\cdot||$ over the image of the Riemannian metric $\tilde{g}$ in order to, for every $\varepsilon > 0$, there exists $\delta(\tilde{g}) > 0$ such that
$$q \in B_{\delta(\tilde{g})}(p) \Longrightarrow |\tilde{g}_{ij}(q) - \tilde{g}_{ij}(p)| \leq = ||\tilde{g}(q) - \tilde{g}(p)|| < \varepsilon$$
Choosing $C > 1$ and $\varepsilon := \frac{1}{2} \left( C - \frac{1}{C} \right)$, we have
$$\frac{1}{C} \delta_j^i \leq \tilde{g}_{ij} \leq C \delta_j^i \ (1)$$
on $B_{\delta(\tilde{g})}(p)$ for each $p \in M$.
I do not sure how to do this, once that $\mathscr{L}^2(T_pM \times T_pM, \mathbb{R})$ and the coordinate fields vary with $p$, therefore I think I can not take simply the operator norm of this space to be $||\cdot||$, but if I can overcome this difficult, then we can do an analogous reasoning for $g$ to obtain
$$\frac{1}{C} \delta_j^i \leq g_{ij} \leq C \delta_j^i \ (2)$$
on $B_{\delta(g)}(p)$ for each $p \in M$.
Defining $\delta := \min \{ \delta(\tilde{g}), \delta(g) \}$, $(1)$ and $(2)$ hold on $B_{\delta}(p)$ for each $p \in M$. Combining $(1)$ and $(2)$ and observing that $\{ B_{\delta}(p) \ ; \ p \in M \}$ is an cover for $M$, we proved the inequalities desired.
$\textbf{EDIT:}$
We know that
$$\frac{1}{A} g_p(v,v) \leq \tilde{g}_p(v,v) \leq A g_p(v,v) \ (\star)$$
for all $v \in T_pM$ based on what DIdier_ proved. Analogously,
$$\frac{1}{B} \tilde{g}_p(v,v) \leq g_p(v,v) \leq B \tilde{g}_p(v,v) \ (\star \star)$$
for all $v \in T_pM$.
I will try to prove that
$$\frac{1}{C} g_p(u,v) \leq \tilde{g}_p(u,v) \leq C g_p(u,v)$$
for all $u,v \in T_pM$.
Let $q_{g_p}(v) := g_p(v,v)$ and $q_{\tilde{g}_p}(v) := \tilde{g}_p(v,v)$ be the quadratic forms associated to the $g_p$ and $\tilde{g}_p$ respectively, then
$$g_p(u,v) = \frac{q_{g_p}(u+v) - q_{g_p}(u) - q_{g_p}(v)}{2} \ \text{and} \ \tilde{g}_p(u,v) = \frac{q_{\tilde{g}_p}(u+v) - q_{\tilde{g}_p}(u) - q_{\tilde{g}_p}(v)}{2}.$$
This, $(\star)$ and $(\star \star)$ imply that
$$\tilde{g}_p(u,v) \leq \left( A - \frac{1}{A} \right) g_p(u,v)$$
and
$$g_p(u,v) \leq \left( B - \frac{1}{B} \right) \tilde{g}_p(u,v)$$
for all $u,v \in T_pM$, therefore
$$\frac{1}{\left( B - \frac{1}{B} \right)} g_p(u,v) \leq \tilde{g}_p(u,v) \leq \left( A - \frac{1}{A} \right) g_p(u,v)$$
for all $u,v \in T_pM$.
Choosing $C > 1$ sufficiently large such that
$$\frac{1}{C} g_p(u,v) \leq \frac{1}{\left( B - \frac{1}{B} \right)} g_p(u,v) \leq \tilde{g}_p(u,v) \leq \left( A - \frac{1}{A} \right) g_p(u,v) \leq C g_p(u,v)$$
for all $u,v \in T_pM$ gives the result.
 A: The point of this answer is to explain what the question is; the other answer has a perfect proof.
As stated in your first quote, $\frac{1}{C} g \leq \tilde{g} \leq C g$ is to be understood in the sense of quadratic forms.
This means that for all $x\in M$ and $v\in T_xM$ we have
$$
\frac{1}{C} g_x(v,v) \leq \tilde{g}_x(v,v) \leq C g_x(v,v)
$$
or equivalently in local coordinates
$$
\frac{1}{C} \sum_{i,j}g_{ij}(x)v^iv^j \leq \sum_{i,j}\tilde{g}_{ij}(x)v^iv^j \leq C \sum_{i,j}g_{ij}(x)v^iv^j.
$$
This means that the two norms on every $T_xM$ are bi-Lipschitz equivalent and the constant is independent of $x$.
Even when we write $\frac{1}{C} g_{ij} \leq \tilde{g}_{ij} \leq C g_{ij}$, it can be a shorthand for the inequalities in the sense of quadratic forms.
That is indeed a much more likely interpretation than a componentwise result.
To underline the importance of working with quadratic forms rather than individual components, let me define three (partial) orders for symmetric square matrices:

*

*In the sense of quadratic forms: $A\leq_{qf}B$ means that $v^TAv\leq v^TBv$ for all $v$.

*Componentwise: $A\leq_{cw}B$ means that $A_{ij}\leq B_{ij}$ for all indices.

*For all pairs: $A\leq_{p}B$ means that $u^TAv\leq u^TBv$ for all $u$ and $v$.

Now take
$$
A
=
\begin{pmatrix}
1&0\\
0&1
\end{pmatrix}
$$
and
$$
B
=
\begin{pmatrix}
1&10\\
10&1
\end{pmatrix}.
$$
Clearly $A\leq_{cw}B$, but for $v=(1,-1)$ we have
$$
2
=
v^TAv
>
v^TBv
=
-18.
$$
Thus $A\leq_{cw}B$ does not imply $A\leq_{qf}B$.
In the case of Riemannian metrics proving $\frac1Cg\leq_{cw}\tilde g\leq_{cw}Cg$ is insufficient, and in general it does not even hold.
For example, if $\tilde g$ is the Euclidean metric (the identity matrix) and $g$ is a Riemannian metric with non-zero (perhaps both positive and negative) off-diagonal entries at some point, the componentwise version is false but the version with quadratic forms is still valid.
In general, $A\leq_{p}B$ implies both $A\leq_{qf}B$ (use the same vector twice) and $A\leq_{cw}B$ (choose two basis vectors).
While the order given by pairs of vectors implies the correct one, it often fails because because the componentwise one does even though the desired estimate holds true.
What you need is $\frac1Cg\leq_{qf}\tilde g\leq_{qf}Cg$, not $\frac1Cg\leq_{cw}\tilde g\leq_{cw}Cg$ or $\frac1Cg\leq_{p}\tilde g\leq_{p}Cg$.
Unfortunately your proof that $\frac1Cg\leq_{qf}\tilde g\leq_{qf}Cg$ implies $\frac1Cg\leq_{cw}\tilde g\leq_{cw}Cg$ is invalid.
A: 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 $TM$.
