带有绝对值的不等式

hbghlyj

类似题
\[\frac{|x-y|}{1+|x|+|y|}\leq\frac{|x-z|}{1+|x|+|z|}+\frac{|z-y|}{1+|z|+|y|}\quad\text{for all $x,y,z∈ℝ$}\]

hbghlyj

证明必须要用到$\Bbb R$的性质。
因为对于一般的度量$d$，$d_1(x,y)=\frac{d(x,y)}{1+d(0,x)+d(0,y)}$不一定满足三角不等式。
反例：for $\mathbb R^2$ with $\ell_\infty$ norm: $x= (2,0)$, $y=(-1,-1)$, $z=(0, -2)$. Here $\nu(x,y)=0.75$ while $\nu(x,z)=0.4$ and $\nu(y,z) = 0.25$. So there can't be an elegant argument for general metric space... But I don't have any counterexamples for Euclidean norm. Of course, for Euclidean norm it suffices to deal with $\mathbb R^3$.

Aluminiumor

好久没做出来，顶一下

hbghlyj

Aluminiumor 发表于 2025-4-16 13:25
好久没做出来，顶一下

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