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[不等式] 双曲直角三角形斜边$>\sqrt{a^2+b^2}$

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hbghlyj Posted at 2023-8-22 00:53:29 |Read mode
Last edited by hbghlyj at 2024-4-1 15:29:00trigonometry of hyperbolic right triangles
$a,b,c\inR,\cosh c=\cosh a\cosh b,$ 求证$c^2>a^2+b^2$

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 Author| hbghlyj Posted at 2023-8-22 03:33:53

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Czhang271828 Posted at 2023-8-22 15:34:14
$\ln \cosh \sqrt x=\ln \cosh \sqrt y+\ln \cosh \sqrt z$, 求证 $x>y+z$.

仅考虑 $\mathbb R_+$. 注意到 $f(x):=\ln \cosh \sqrt x$ 的一阶导为 $f'(x)=\dfrac{\tanh \sqrt x}{2\sqrt x}>0$, 显然 $f''(x)<0$. 现已知 $f(x)-f(y)=f(z)-f(0)$, 以及 $y>0$, 从而 $x-y>z-0$.

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 Author| hbghlyj Posted at 2023-8-22 17:35:32
Czhang271828 发表于 2023-8-22 15:34
$\ln \cosh \sqrt x=\ln \cosh \sqrt y+\ln \cosh \sqrt z$, 求证 $x>y+z$.

仅考虑 $\mathbb R_+$. 注意到 $f(x):=\ln \cosh \sqrt x$ 的一阶导为 $f'(x)=\dfrac{\tanh \sqrt x}{2\sqrt x}>0$, 显然 $f''(x)<0$. 现已知 $f(x)-f(y)=f(z)-f(0)$, 以及 $y>0$, 从而 $x-y>z-0$.
🙂类似于我在那个帖子下的回答

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