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[不等式] 一道不等式小菜

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facebooker Posted 2021-1-26 17:55 |Read mode

pro:$\sqrt{x^{2}-1}+\sqrt{x-1}\leqslant x\sqrt{x}$

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kuing Posted 2021-1-26 20:19

懒得想，直接平方移项再平方，最后分解为 `x^2(x^2-x-1)^2\geqslant0`，看来取等和黄金分割有关……

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