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Dmitri Panov
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This inequality is not true. Consider the rectangle on $\mathbb R^2$ with vertices $(\pm 1,\pm \varepsilon)$$(\pm 1, 0)$, $(0, \pm \varepsilon)$. Then on the left you have $2\varepsilon(4-8/\pi)$ on the right you have approximatively $4\varepsilon^2$.

This inequality is not true. Consider the rectangle on $\mathbb R^2$ with vertices $(\pm 1,\pm \varepsilon)$. Then on the left you have $2\varepsilon(4-8/\pi)$ on the right you have approximatively $4\varepsilon^2$.

This inequality is not true. Consider the rectangle on $\mathbb R^2$ with vertices $(\pm 1, 0)$, $(0, \pm \varepsilon)$. Then on the left you have $2\varepsilon(4-8/\pi)$ on the right you have approximatively $4\varepsilon^2$.

Source Link
Dmitri Panov
  • 28.9k
  • 4
  • 92
  • 161

This inequality is not true. Consider the rectangle on $\mathbb R^2$ with vertices $(\pm 1,\pm \varepsilon)$. Then on the left you have $2\varepsilon(4-8/\pi)$ on the right you have approximatively $4\varepsilon^2$.