The inequality can't be true for all M. I agree with Willie Wong's approach and conclusion: the question reduces to whether the tangent to MB at v $v$ is orthogonal to v. $v$. Note that, since $v$ has minimum $l^{3/2}$ norm among all points in the image of the unit $l^3$ ball MB, then the tangent to MB at v $v$ is also tangent to the $l^{3/2}$ sphere of radius ||v||3/2, and it is orthogonal to v, $v$, only in the special case of v $v$ in the axes or in the diagonals (in ${\mathbb R}^3$, there are 26 such special points) So, it is sufficient to consider an example of M where v $v$ is not one of these special points. In fact, it's quite clear that M can be chosen so that the minimizing point v $v$ is any prescribed point, by a continuity argument. Fixing $m_{3,1}=m_{3,2}=0$ makes everything 2 dimensional, and even easier.
 Example with $m_{3,1}=m_{3,2}=0$ (so that one can consider M as a 2x2 matrix). Consider for instance a rotation $M=M_\theta$ of a small angle $\theta$. As above, let $x$ be any point minimizing $\|Mx\|_{3/2}$ on the unit sphere $S:=\{\xi\ : \|\xi\|_3=1\}$. Since $\theta$ is small, $Mx$ has to be close to the corresponding minimizers with $\theta=0$, that is, to one of the 4 unit points in the axes; actually it will be enough to know that $Mx$ is not one of the 4 points on the diagonals. Also, for small $\theta\neq0$, $Mx$ can't be in the axes either (the tangent to $MS$ in a point $v$ which is in the axes is not orthogonal to the axis: therefore $v\neq Mx.$) Since the only points of a $l^{3/2}$ sphere where the tangent is orthogonal to the radius are either on the axes or on the diagonals, one has that the tangent to $MS$ at $Mx$, which by minimality coincides with the tangent to the $l^{3/2}$ sphere at $Mx$, is not orthogonal to $Mx$; this implies that the projection of $MB$ onto the line spanned by $v=Mx$ covers $v$ itself, as already observed in WW's answer.
The inequality can't be true for all M. I agree with Willie Wong's approach and conclusion: the question reduces to whether the tangent to MB at v is orthogonal to v. Note that, since $v$ has minimum $l^{3/2}$ norm among all points in the image of the unit $l^3$ ball MB, then the tangent to MB at v is also tangent to the $l^{3/2}$ sphere of radius ||v||3/2, and it is orthogonal to v, only in the special case of v in the axes or in the diagonals (in ${\mathbb R}^3$, there are 26 such special points) So, it is sufficient to consider an example of M where v is not one of these special points. In fact, it's quite clear that M can be chosen so that the minimizing point v is any prescribed point, by a continuity argument. Fixing $m_{3,1}=m_{3,2}=0$ makes everything 2 dimensional, and even easier.
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