$\sup \left\| A x + B y + C z \right\|$ subject to $\left\|x\right\| = \left\|y\right\| = \left\|z\right\| = 1$ I'm interested in finding

$\sup \left\| A x  + B y + C z \right\|$ subject to 
$\left\|x\right\| = \left\|y\right\| = \left\|z\right\| = 1$

where $A$, $B$, $C$ and $x$, $y$, $z$ are real matrices and vectors, respectively, of compatible sizes, and the norms are Euclidean.

What is this problem called in the literature? (If there is no established name, does this problem reduce to a well-known one?)

(If there was only one constraint, this would be the induced matrix norm)



Refutation of @zimbra314's answer (Can not fit in the comments)


  $[x, y, z]^{T}= \alpha_{1} v^{T}_{1} + \alpha_{2} v^{T}_{2}+\alpha_{3} v^{T}_{3}$, where $v_{i}$ is the $i$-the right singular vector of $[A,B,C]$ (i.e. singular vector corresponding to $i$-th largest singular value).


This assumes that the optimal $[x, y, z]^{T}$ is a linear combination of the first 3 right singular vectors of $[A, B, C]$. However, this assumption is incorrect. As a counterexample, consider

$
A = \begin{bmatrix}2&amp;0&amp;0\\0&amp;2&amp;0\\0&amp;0&amp;2\\0&amp;0&amp;0\\0&amp;0&amp;0\end{bmatrix}
$ 

$
B = \begin{bmatrix}0\\0\\0\\1\\0\end{bmatrix}
$ 

$
C = \begin{bmatrix}0\\0\\0\\0\\1\end{bmatrix}
$
Hint: 

$\left \| Ax+By+Cz \right \| = \left \| \left [ A, B, C \right ] \left [ \begin{matrix}
x\\ 
y\\ 
z
\end{matrix} \right ]\right \|$

Detailed answer: 

$[x, y, z]^{T}= \alpha_{1} v^{T}_{1} + \alpha_{2} v^{T}_{2}+\alpha_{3} v^{T}_{3}$, where $v_{i}$ is the $i$-the right singular vector of $[A,B,C]$ (i.e. singular vector corresponding to $i$-th largest singular value).  Denoting $[v_{ix},v_{iy},v_{iz}]^{T}=v_{i}$ to mark the partition of $v_{i}$ corresponding to $x$, $y$ and $z$.

Now solve: $||x||^{2}=1 \rightarrow ||(\alpha_{1} v_{1x} + \alpha_{2} v_{2x}+\alpha_{3} v_{3x})||=1$ 

Similarly ,
$||(\alpha_{1} v_{1y} + \alpha_{2} v_{2y}+\alpha_{3} v_{3y})||=1$ 

$||(\alpha_{1} v_{1z} + \alpha_{2} v_{2z}+\alpha_{3} v_{3z})||=1$

Three equations to solve for three unknowns $\alpha_{1}, \alpha_{2}$, and $\alpha_{3}$. Then, your answer is $\sqrt{\alpha^{2}_{1} \sigma^{2}_{1} + \alpha^{2}_{2} \sigma^{2}_{2} + \alpha^{2}_{3} \sigma^{2}_{3}}$, where $\sigma_{i}$ is the singular value corresponding to $i$-the largest singular vectors.
Hint: 

$\left \| Ax+By+Cz \right \| = \left \| \left [ A, B, C \right ] \left [ \begin{matrix}
x\\ 
y\\ 
z
\end{matrix} \right ]\right \|$

Detailed answer: 

$[x, y, z]^{T}= \alpha_{1} v^{T}_{1} + \alpha_{2} v^{T}_{2}+\alpha_{3} v^{T}_{3}$, where $v_{i}$ is the $i$-the right singular vector of $[A,B,C]$ (i.e. singular vector corresponding to $i$-th largest singular value).  Denoting $[v_{ix},v_{iy},v_{iz}]^{T}=v_{i}$ to mark the partition of $v_{i}$ corresponding to $x$, $y$ and $z$.

Now solve: $||x||^{2}=1 \rightarrow ||(\alpha_{1} v_{1x} + \alpha_{2} v_{2x}+\alpha_{3} v_{3x})||=1$ 

Similarly ,
$||(\alpha_{1} v_{1y} + \alpha_{2} v_{2y}+\alpha_{3} v_{3y})||=1$ 

$||(\alpha_{1} v_{1z} + \alpha_{2} v_{2z}+\alpha_{3} v_{3z})||=1$

Three equations to solve for three unknowns $\alpha_{1}, \alpha_{2}$, and $\alpha_{3}$. Then, your answer is $\sqrt{\alpha^{2}_{1} \sigma^{2}_{1} + \alpha^{2}_{2} \sigma^{2}_{2} + \alpha^{2}_{3} \sigma^{2}_{3}}$, where $\sigma_{i}$ is the singular value corresponding to $i$-the largest singular vectors.