[不等式] $x+y+z=1$，求证：$\sum\dfrac{x}{\sqrt{y+z}}\geqslant\dfrac{\sqrt6}2$

已知$x、y、z\in R^+$，且$x+y+z=1$，求证：$\sum\dfrac{x}{\sqrt{y+z}}\geqslant\dfrac{\sqrt6}2$

1# yes94
$f(x)=\frac{x}{\sqrt{1-x}}$是下凸函数,用琴生不等式或$x=\frac{1}{3}$处切线法都可以的,也许cauchy不等式也行.

2# realnumber
不错，切线法可以搞：

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2013-2-15 11:56

There also  exist a proof with Holder inequality :D

There also  exist a proof with Holder inequality :D
pxchg1200 发表于 2013-2-15 18:02

那我来试一试：

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2013-2-15 18:39

是不是这个？如果不是请给出解答，谢谢！

5# yes94


There are many ways to apply Holder inequality :D
\[ \left(\sum{\frac{x}{\sqrt{y+z}}} \right)^{2}\left(\sum{x(y+z)} \right)\geq (x+y+z)^{3} \]

The result is easy :D