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[不等式] 一道四元不等式

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longzaifei Posted at 2014-3-5 08:19:31 |Read mode
$x,y,z,w\in[-1,1] ,x+y+z+w=0,$   证明:\[  \sqrt{1+x+y^2}+\sqrt{1+y+z^2}+\sqrt{1+z+w^2}+\sqrt{1+w+x^2}\ge 4    \]

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tommywong Posted at 2014-3-5 09:01:43
琴生不等式

$\sqrt{1+x+y^2}+\sqrt{1+y+z^2}+\sqrt{1+z+w^2}+\sqrt{1+w+x^2} \ge 4\sqrt{1+\frac{x+y+z+w}{4}+\frac{x^2+y^2+z^2+w^2}{4}} \ge 4\sqrt{1+\frac{x+y+z+w}{4}+(\frac{x+y+z+w}{4})^2} = 4$

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 Author| longzaifei Posted at 2014-3-5 10:20:56
设$f(x)=\sqrt{1+x} $, 但是 $f(x) $是上凸函数,第一个不等号应该是 $\le $

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tommywong Posted at 2014-3-7 18:47:25
似乎要用到这题的方法
kuing.cjhb.site/forum.php?mod=viewthread& … e%26amp%3Btypeid%3D1

但也可能是把x,y,z,w换成三角函数

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