Let $(M,g)$ be a Riemannian manifold. Let $S\_g$ be the corresponding Sasaki metric on $TM$. For every $p\in M$, $V\_p\in T\_pM$, is it true and obvious that $0\_p$ is the closest point of the zero section to $V\_p$?


With some abuse of terminology a rephrase of the question would be: Is the height of a right triangle shorter than its hypotenuse?



The distance to the zero section of $TM$
*Let me expand my comment to remove the question from unanswered.*


Any point in $w\in\mathrm{T}M$ is a pair $w=(V,p)$ where $p\in M$ and $V\in\mathrm{T}\_pM$.


Let $t\mapsto w(t)=(V(t),\gamma(t))$ be a curve in $\mathrm{T}M$.
Let $V=V(0)$ and $p=\gamma(0)$.
Note that
$$w'(0)=\nabla\_{\gamma'(0)}V\oplus \gamma'(0)\in \mathrm{T}\_{p}M\oplus\mathrm{T}\_{p}M=\mathrm{T}\_{(V,p)}\mathrm{T}M.$$


Therefore
$$
\begin{aligned}
\langle V\oplus 0,w'(0)\rangle
&=\langle V,\nabla\_{\gamma'(0)}V\rangle=
\\
&=\tfrac12\cdot\langle V,V\rangle'(0).
\end{aligned}
$$
It follows that $(V,p)\mapsto V\oplus 0$ is the gradient of the function $f\colon(V,p)\mapsto \tfrac12\cdot\langle V,V\rangle$.
Whence the statement follows.


