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Judging whether one convex polytope is inside of another when both are expressed as a system of linear inequalities seems to be a complex problem. This answer 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?

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 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?

This answer claims the following proposition.

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?

Judging whether one convex polytope is inside of another when both are expressed as a system of linear inequalities seems to be a complex problem. This answer 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?