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[不等式] 请教一个不等式问题

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snowblink posted 2024-1-30 00:04 |Read mode
已知正数$a$,$b$,$c$满足$a+b+c=1$,求证:$\dfrac{\sqrt{a} }{1-a}+ \dfrac{\sqrt{b} }{1-b}+\dfrac{\sqrt{c} }{1-c}\geq \dfrac{3\sqrt{3} }{2} $.

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爪机专用 posted 2024-1-30 01:08 from mobile
分子分母同乘分子,将1-a移入根号内,然后用三元均值

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昨晚在床上用手机回不方便打代码,补充如下:  posted 2024-1-30 14:25
\[\frac{\sqrt a}{1-a}=\frac a{\sqrt a(1-a)}=\frac{\sqrt2a}{\sqrt{2a(1-a)(1-a)}}\ge\frac{\sqrt2a}{\sqrt{\left( \frac{2a+1-a+1-a}3 \right)^3}}=\frac{3\sqrt3}2a\]  posted 2024-1-30 14:25
I am majia of kuing

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ic_Mivoya posted 2024-1-30 12:25
只需注意到 $0<x<1$ 时 $\dfrac{\sqrt x}{1-x}\ge \dfrac{3\sqrt3}2x$.

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