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[不等式] 如何证明三角形中两个大小关系?

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lemondian Post time 2024-4-28 11:00 |Read mode
在三角形中,有$s^2\leqslant \dfrac{R(4R+r)^2}{2(2R-r)}$,$s^2\leqslant 2R^2+10Rr-r^2+2(R-2r)\sqrt{R^2-2Rr}$,
如何证明:$2R^2+10Rr-r^2+2(R-2r)\sqrt{R^2-2Rr}\leqslant \dfrac{R(4R+r)^2}{2(2R-r)}$呢?

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kuing Post time 2024-4-28 12:43
\[2\sqrt{R^2-2Rr}\leqslant\frac{2R-r}2+\frac{2(R^2-2Rr)}{2R-r},\]
代入即得

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收到,谢谢!  Post time 2024-4-29 09:36

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