Endpoint-average inequality for a line segment in a normed space Let $X$ be a normed vector space   over $\mathbb R$. What is the smallest universal constant $C>0$ such that the inequality
$$\|x\|\le C\int_0^1 \|x+tv\|\,dt\tag{1}$$
holds for all $x,v\in X$? 
Geometrically, (1) means that the norm of an endpoint of a line segment can be majorized by the average of the norm over said line segment.
More general version: for $1\le p<\infty$, what is the smallest $C_p$  such that 
$$\|x\|^p\le C_p\int_0^1 \|x+tv\|^p\,dt\tag{2}$$
holds for all $x,v\in X$? 

Partial results: in this answer I show $C_2\le 27$ by a  clunky estimate that is certainly not sharp. Applied to general $p$, the same estimate yields $C_p\le 3^{p+1}$. 
By taking $v=-x$, one finds a lower estimate: $C_p\ge p+1$. Which is not sharp, because $v=-(1+\epsilon)x$ will outperform this.
One can take $X$ to be 2-dimensional: everything happens in the  span of $x$ and $v$.
 A: For $x = 0$, any $C$ will do, hence we can by homogeneity assume $\lVert x\rVert = 1$. So we want to find
$$\frac{1}{C_p} = \inf \left\lbrace \int_0^1 \lVert x+tv\rVert^p\,dt : v \in X\right\rbrace.$$
We can also assume $v\neq 0$, since $v = -x$ yields a smaller result than $v = 0$. By the triangle inequality, we have
$$\begin{align}
\lVert x+tv\rVert &\geqslant 1 - t\lVert v\rVert,\quad 0 \leqslant t \leqslant \frac{1}{\lVert v\rVert},\\
\lVert x+tv\rVert &\geqslant t\lVert v\rVert - 1,\quad \frac{1}{\lVert v\rVert} \leqslant t.
\end{align}$$
We have equality if $v$ is a negative multiple of $x$. For $\lVert v\rVert \leqslant 1$, it is clear that
$$\int_0^1 \left(1-t\lVert v\rVert\right)^p\,dt$$
is decreasing in $\lVert v\rVert$, and for $\lVert v\rVert = 1$ the value is $\dfrac{1}{p+1}$. For $r = \lVert v\rVert > 1$, we want to find the minimum of
$$\begin{align}
I_p(r) &= \int_0^{1/r} (1-rt)^p\,dt + \int_{1/r}^1 (rt-1)^p\,dt\\
&= \frac{1}{(p+1)r} + \frac{(r-1)^{p+1}}{(p+1)r}.
\end{align}$$
We have $$I_p'(r) = \frac{(p+1)r(r-1)^p - (r-1)^{p+1} - 1}{(p+1)r^2} = \frac{(r-1)^p(pr+1)-1}{(p+1)r^2},$$
so $I_p'(1)  < 0$. Further, it is clear that $\lim\limits_{r\to\infty} I_p(r) = +\infty$, so the unique critical point of $I_p$ is where the global minimum is attained. Also, $I_p'(2) = \dfrac{p}{2(p+1)} > 0$, therefore the critical point lies between $1$ and $2$. That also follows from $I_p(2) = I_p(1)$.
I can't solve the general critical point equation
$$(pr+1)(r-1)^p = 1$$
exactly, but for $p = 1$, the value is $r = \sqrt{2}$, with $I_1(\sqrt{2}) = \sqrt{2}-1$, so
$$C_1 = \sqrt{2}+1.$$
For $p = 2$, the critical point is $r = \frac32$, with $I_2\left(\frac32\right) = \frac14$, so
$$C_2 = 4.$$
For large $p$, the critical point $r_p$ is approximately
$$r_p^{(0)} = 2 - \frac{\log p + \log 2}{p}.$$
Anyway, we have $(r_p-1)^p = \dfrac{1}{pr_p+1}$, and thus
$$I_p(r_p) = \frac{1+(r_p-1)^{p+1}}{(p+1)r_p} = \frac{1 + \frac{r_p-1}{pr_p+1}}{(p+1)r_p} = \frac{1}{pr_p+1},$$
which means
$$C_p = pr_p + 1 < 2p+1,$$
and the bound $2p+1$ is not too bad [asymptotically, $C_p = 2p + O(\log p)$] (however, the approximation $r_p^{(0)}$ suggests that $2p + 1 - \log p - \log 2$ is a better approximation to $C_p$ for large $p$).
