norm inequalities Let $p>2$. I'd like to know the best possible lower and upper bound for $\|x\|_p$ given that $x\in R^n$ and $\|x\|_1$, $\|x\|_2$, and $\|x\|_\infty$ have fixed values.
It is well-known that 
$$\|x\|_\infty\le \|x\|_p\le \|x\|_2^{2/p}\|x\|_\infty^{1-2/p}~~~
\Big[\le \|x\|_2\le \|x\|_1\Big].$$
But these bounds cannot be sharp when an arbitrary value of $\|x\|_1$ is specified. 
In case no closed form solution is known, I'd also appreciate pointers to explicit suboptimal bounds that are stronger than the stated ones. For example, I conjecture that a bound of the form
$$ \|x\|_p\le C_p\|x\|_2\Big(\frac{\|x\|_\infty}{\|x\|_1}\Big)^r$$
should hold with $r=(p-2)/(2p-2)$ and a constant $C_p$ depending on $p$ only, but I don't have a proof of such an inequality.
 A: A likely candidate for the optimal solution is when the $x_i$ take $3$ different values, say $x_i = a$ for $i=1 \ldots n_1$, $b$ for $i=n_1+1 \ldots n_1 + n_2$, $\|x\|_\infty$ for $i=n_1+n_2+1 \ldots n$, where $n_i \ge 0$, $ n_1 + n_2 \le n$, and $a$ and $b$ must satisfy
$$ \eqalign{ n_1 a + n_2 b + (n-n_1-n_2) \|x\|_\infty &= \|x\|_1\cr
             n_1 a^2 + n_2 b^2 + (n-n_1-n_2) \|x\|_\infty^2 &= \|x\|_2^2\cr
             0 \le a,b &\le \|x\|_\infty
            } $$
For given $n, n_1, n_2, \|x\|_1, \|x\|_2, \|x\|_\infty$, this will determine $a$ and $b$ (up to interchange) as the roots of a quadratic, when that quadratic has 
two roots in the interval $[0, \|x\|_\infty]$. Which $n_1$ and $n_2$ give the maximum or minimum value of $\|x\|_p$ will likely vary. 
A: This answers the question of Iosif that was asked in comments (see comments under OP). 
Theorem Let $\gamma(t)=(t,f(t),g(t)) : [0,1] \to \mathbb{R}^{3}$ be a space curve such that  $f''>0$, and the torsion of $\gamma$ is positive, then the surface parametrized as $P(s,t)=s\gamma(t)+(1-s)\gamma(1)$, $(s,t) \in [0,1]^{2}$ represents the upper boundary of the convex hull of the curve $\gamma$. 
Since the surface consists of the line segments $[\gamma(t),\gamma(1)]$, $t \in [0,1]$; and $P(1,t)=\gamma(t)$,  therefore it is enough to show that the surface is concave. To prove the latter fact one may try to compute the first and the second fundamental forms of a surface given in a parametric way. One obtains certain long expressions and it is not clear to me right now how to say something about their signs, however, I would be interested to see if one can push this path until the end.
The proof of the concavity of $P(s,t)$ that I knew involved plenty of computations in the spirit of Section 3 (see the reference below). Here I will present a short argument of Pavel Zatitskiy.
Proof:
We need to prove that for any point $t_{0}\in [0,1)$, one can draw a supporting hyperplane $T$ to the surface $P$ such that $T$ contains the segment  $[\gamma(t_{0}),\gamma(1)]$. Let us subtract  from $g$ a linear combination of $f(t),t,1$ so that the new function $h(t)=g(t)-c_{1}f(t)-c_{2}t-c_{3}$ would satisfy the properties: $h(t_{0})=h'(t_{0})=h(1)=0$. Such linear combination exists for example beacuse the vectors $(1,f'(t_{0})), (f(t_{0})-f(1),t_{0}-1)$ are linearly independent. So what we did so far is that we translated and rotated the picture so that our hypothetical tangent plane $T$ to be horizontal just for convenience. Thus it is enough to show that $h\leq 0$ on $[0,1]$.
Next, positivity of the torsion of $\gamma$ means that $f''g'''-f'''g''>0$, i.e., $g''/f''$ is increasing. This in partiuclar means that $h''/f''$ is increasing. Since $f''>0$ this implies that $h''$  changes sign from - to + at most once, i.e., $h$ is concave and then convex. We need to check that the point $t_{0}$ lies in the domain where $h$ is concave, i.e., $h''(t_{0})<0$ (this should follow from the initial conditions on $h$). Then we obtain that $h$ is concave on a certain segment $[0,t_{1}]$, and then convex on $[t_{1},1]$, in addition $t_{0}<t_{1}$. Using the facts that $h(t_{0})=h'(t_{0})=h(1)=0$  we obtain the inequality $h\leq 0$ on $[0,1]$. $\square$ 
Remark 1: For the curve $\gamma(t)=(t,t^{2},t^{p})$, $t\in [0,A]$, since $\tau_{\gamma}>0$, the  theorem immediately gives the concavity of $P(s,t)$, which explicity can be rewritten as the graph of the function 
$$
B(x,y)=\frac{(y-x^{2}) A^{p}+(Ax-y)^{p}(A-x)^{2-p}}{(A-x)^{2}+y-x^{2}}, \quad p\geq 2,
$$
in the domain $\Omega_{A} = \{(x,y)\, :\, Ax \geq y \geq x^{2},\, x\geq 0\}$ with the boundary condition $B(t,t^{2})=t^{p}$. This gives the sharp upper bound in OP. Indeed, take any $z=(z_{1},\ldots, z_{n})$ with $\|z\|_{\infty}\leq A$. WLOG $z_{j}\geq 0$. Let $\|z\|^{q}_{q}:=(1/n)\sum_{j=1}^{n}z^{q}$, then 
$$
\|z\|_{p}^{p}=\frac{1}{n}\sum_{j=1}^{n}z_{j}^{p} = \frac{1}{n}\sum_{j=1}^{n}B(z_{j},z_{j}^{2})\leq B(\|z\|_{1}, \|z\|_{2}^{2}) =\frac{(\|z\|_{2}^{2}-\|z\|_{1}^{2})\|z\|_{\infty}^{p}+(\|z\|_{\infty}\|z\|_{1}-\|z\|_{2}^{2})^{p}(\|z\|_{\infty}-\|z\|_{1})^{2-p}}{(\|z\|_{\infty}-\|z\|_{1})^{2}+\|z\|_{2}^{2}-\|z\|_{1}^{2}}.
$$ 
Remark 2: The theorem, in particular, gives the positive answer to When is the convex hull of two space curves the union of lines? in the case of a curve discussed in the theorem. We verified the upper boundary. The lower boundary of the convex hull will be $s\gamma(0)+(1-s)\gamma(t)$, $[s,t]\in [0,1]^{2}$ by the same reasons. And then it is not difficult to see that everythign between upper and lower boundary can be filled out also by the stright line segments
Ivanisvili, Paata, Inequality for Burkholder’s martingale transform, Anal. PDE 8, No. 4, 765-806 (2015). ZBL1341.60031. 
