Judging whether one convex polytope is inside of another when both are expressed as a system of linear inequalities seems not to be an easy problem.
[This answer](https://math.stackexchange.com/a/198091/64809) on math.stackexchange.com claims the following proposition regarding a special case.


Using Farkas' lemma to show, a halfspace $\{x \mid a' x \leq b' \}$ contains the polytope $P= \{x |A x \leq b \}$ if and only if there exists $\lambda \in \mathbb{R}^m$ with $\lambda \geq 0$ (understood component-wise) such that:
\begin{align\*}
a' &= \lambda^T A \\
b' &\geq \lambda^T b
\end{align\*}


It is easy to prove the sufficiency. I am able to show the necessity of $a' = \lambda^T A$ by Farkas' lemma, but not $b' \geq \lambda^T b$. I tried but failed to argue by contradiction. How does one prove that last inequality?



Confining a polytope to one side of an affine hyperplane
Here is a proof that uses directly the [separating hyperplane theorem](https://en.wikipedia.org/wiki/Hyperplane_separation_theorem), which states the following.


Given a matrix $A$ and column matrix $b$,
\begin{align}
Ax=b, &\text{ for some column matrix }x\ge0 \\
&\iff \\
y^TA\ge0 &\text{ for some column matrix }y \implies y^Tb\ge0
\end{align}




---


We put as follows the required proposition into the above form with the exception that the matrix is transposed. The premise of the proposition is
\begin{align}
&\begin{bmatrix}
A & b \\
0 & 1
\end{bmatrix}\begin{bmatrix}
-xx\_1 \\
x\_1
\end{bmatrix}\ge 0 \\
&\iff \\
&\begin{cases} Ax\le b \implies a'^Tx\le b' \\
x\_1\ge0 
\end{cases}
\\
&\implies \\
&\begin{bmatrix}a'^T &b'\end{bmatrix} \begin{bmatrix}-xx\_1 \\x\_1 \end{bmatrix}\ge0
\end{align}


The conclusion of the proposition is
\begin{align}
&\begin{bmatrix}\lambda^T & \lambda\_1\end{bmatrix}\begin{bmatrix}
A & b \\
0 & 1
\end{bmatrix}=\begin{bmatrix}a'^T & b'\end{bmatrix}, \text{ for some} \begin{bmatrix}\lambda^T & \lambda\_1\end{bmatrix}\ge0 \\
&\iff \\
&\exists \lambda\ge0 \ni
\begin{cases}
\lambda^TA=a'^T \\ 
\lambda^Tb\le b'
\end{cases}
\end{align}


The premise and the conclusion are seen to be equivalent in view of the separating hyperplane theorem as stated above. We have now proved the desired proposition.


