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本帖最后由 小fisher 于 2020-8-13 08:51 编辑
设x=tanA, y=tanB(A,B∈(0, π/2))
\((\frac{\sqrt{1+x^2}-1}y+1)(\frac{\sqrt{1+y^2}-1}x+1)\\=(\frac{\sqrt{1+\tan^2A}-1}{\tan B}+1)(\frac{\sqrt{1+\tan^2B}-1}{\tan A}+1)\\=(\frac{\displaystyle\frac1{\cos A}-1}{\tan B}+1)(\frac{\displaystyle\frac1{\cos B}-1}{\tan A}+1)\\=(\frac{\displaystyle\frac{1-\cos A}{\cos A}}{\tan B}+1)(\frac{\displaystyle\frac{1-\cos B}{\cos B}}{\tan A}+1)\\=(\frac{\displaystyle\frac{1-\cos A}{\sin A}\cdot\frac{\sin A}{\cos A}}{\tan B}+1)(\frac{\displaystyle\frac{1-\cos B}{\sin B}\cdot\frac{\sin B}{\cos B}}{\tan A}+1)\\=(\frac{\tan A}{\tan B}\cdot\tan\frac A2+1)(\frac{\tan B}{\tan A}\cdot\tan\frac B2+1)\\=\tan\frac A2\cdot\tan\frac B2+\frac{\tan A}{\tan B}\cdot\tan\frac A2+\frac{\tan B}{\tan A}\cdot\tan\frac B2+1\\=2\\\geq\tan\frac A2\cdot\tan\frac B2+2\sqrt{\frac{\tan A}{\tan B}\cdot\tan\frac A2\cdot\frac{\tan B}{\tan A}\cdot\tan\frac B2}+1\\=\tan\frac A2\cdot\tan\frac B2+2\sqrt{\tan\frac A2\cdot\tan\frac B2}+1\\={(\sqrt{\tan\frac A2\cdot\tan\frac B2}+1)}^2\)
解不等式\({(\sqrt{\tan\frac A2\cdot\tan\frac B2}+1)}^2\leq2\),得:
\(\sqrt{\tan\frac A2\cdot\tan\frac B2}\leq\sqrt2-1\)
因此
\(xy=\tan A\tan B=\frac{2\tan{\displaystyle\frac A2}\cdot2\tan{\displaystyle\frac B2}}{(1-\tan^2{\displaystyle\frac A2})(1-\tan^2{\displaystyle\frac B2})}\\\leq\frac{2\tan{\displaystyle\frac A2}\cdot2\tan{\displaystyle\frac B2}}{{(1-\tan{\displaystyle\frac A2\tan\frac B2})}^2}\;\;\;(\mathrm{因为}{(1-mn)}^2\geq(1-m^2)(1-n^2)\\\leq\frac{4(\sqrt2-1)^2}{\left(1-(\sqrt2-1\right)^2)^2}=1\)
令A=B可使两次≤取等号。
昨天太匆忙,缺少了\(\tan\frac A2\cdot\tan\frac B2\)取最大值时tanA·tanB也取最大值的证明过程,已做补充。 |
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