Vector Sum and Minkowski Sum

\begin{theorem}[$Brunn-Minkowski$ inequality for convex bodies \rom{2}] \label{thm:2} Let $K$ and $L$ be two convex bodies (i.e. compact convex sets with nonempty interiors) in $\mathbb{R}^n$, then: \begin{equation} \label{eq:2} V(K+L)^{1/n} \geq V(K)^{1/n} +V(L)^{1/n} \end{equation} \end{theorem} In geometry, the term $Minkowski$ $sum$ is more frequently used for the vector sum. We now give an example to further clarify this claim.
Consider the origin-symmetric rectangle $H$ of sides $a$ and $b$ and the other origin-symmetric rectangle $S$ of sides $c$ and $d$ and their $Minkowski$ $sum$ shown below. \begin{center} \begin{tikzpicture} \draw [purple, very thick] (0,0) -- node[below, pos=.5]{b}(4,0) -- (4,2) -- (0,2) -- node[left, pos=.5]{a}(0,0) ; \end{tikzpicture} \end{center} \ $$\textbf{+}$$ \begin{center} \begin{tikzpicture} \draw [orange, very thick] (6,-0.5) -- node[below, pos=.5]{d}(8,-0.5) -- (8,3) -- (6,3) -- node[left, pos=.5]{c}(6,-0.5) ; \end{tikzpicture} \end{center} $$\textbf{=}$$ \begin{center} \begin{tikzpicture} \draw [orange, very thick] (10,0) -- node[below, pos=.5]{d/2} (11,0); \draw [orange, very thick] (10,0) -- node[left, pos=.5]{c/2} (10,1.75); \draw[dotted] [purple, very thick] (10, 1.75) -- (10, 3.75); \draw [orange, very thick] (10,3.75) -- node[left, pos=.5]{c/2} (10,5.5); \draw [orange, very thick] (10,5.5) -- node[above, pos=.5]{d/2} (11,5.5); \draw[dotted] [purple, very thick] (11, 5.5) -- (15, 5.5); \draw [orange, very thick] (15,5.5) -- node[above, pos=.5]{d/2} (16,5.5); \draw [orange, very thick] (16,5.5) -- node[right, pos=.5]{c/2} (16,3.75); \draw[dotted] [purple, very thick] (16, 3.75) -- (16, 1.75); \draw [orange, very thick] (16,1.75) -- node[right, pos=.5]{c/2} (16,0); \draw [orange, very thick] (16,0) -- node[below, pos=.5]{d/2} (15,0); \draw[dotted] [purple, very thick] (15, 0) -- (11, 0); \draw [orange, very thick] (11,0) -- (11,1.75); \draw [purple, very thick] (11,1.75) -- node [left, pos=.5]{a} (11,3.75); \draw [orange, very thick] (11,3.75) -- (11,5.5); \draw [orange, very thick] (15,0) -- (15,1.75); \draw [purple, very thick] (15,1.75) -- node [right, pos=.5]{a}(15,3.75); \draw [orange, thick] (15,3.75) -- (15,5.5); \draw [orange, very thick] (10,1.75) -- (11,1.75); \draw [purple, very thick] (11,1.75) -- node [below, pos=.5]{b} (15,1.75); \draw [orange, very thick] (15,1.75) -- (16,1.75); \draw [orange, very thick] (10,3.75) -- (11,3.75); \draw [purple, very thick] (11,3.75) -- node [above, pos=.5]{b} (15,3.75); \draw [orange, very thick] (15,3.75) -- (16,3.75); \end{tikzpicture} \end{center}

Now we compute their vector sum to show how it is equivalent to the $Minkowski$ $sum$ $H+S$:

\begin{equation} V(H+S)=V(H)+V(S)+2a(\frac{d}{2})+2b(\frac{c}{2}) \end{equation} \begin{equation} =V(H)+V(S)+ad+bc \end{equation} \begin{equation} \geq V(H)+2\sqrt{V(H)V(S)}+V(S) \end{equation} \begin{equation} =({V(H)}^{\frac{1}{2}}+{V(S)}^\frac{1}{2})^2 \end{equation}

So we get: \begin{equation} V(H+S)^{\frac{1}{2}}\geq V(H)^\frac{1}{2}+V(S)^\frac{1}{2} \end{equation} which is $Minkowski$ inequality ( \ref{eq:2}) for $n=2$. \begin{remark} Notice that we managed to derive equation (8) from equation (7) in the following way: \begin{equation} \begin{split} (ad+bc)^2 &\geq 0\\ (ad)^2+2abcd+(bc)^2 &\geq 0\\ a^2d^2+b^2c^2 &\geq 2abcd\\ a^2d^2+2abcd+b^2c^2 &\geq 4abcd \\ \frac{(ad)^2+2abcd+(bc)^2}{4} &\geq abcd \\ (\frac{ad+bc}{2})^2& \geq V(H)V(S)\\ ad+bc &\geq 2 \sqrt{V(H)V(S)} \end{split} \end{equation} \end{remark}

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