带有绝对值的不等式

APPSYZY

实数 a, b，求证
\[\frac{\abs a}{1+\abs a}+\frac{\abs b}{1+\abs b}\geqslant\frac{\abs{a+b}}{1+\abs{a+b}}.\]

kuing

\begin{align*}
\frac{\abs a}{1+\abs a}+\frac{\abs b}{1+\abs b}&\geqslant\frac{\abs a}{1+\abs a+\abs b}+\frac{\abs b}{1+\abs b+\abs a}\\
&=\frac{\abs a+\abs b}{1+\abs a+\abs b}\\
&\geqslant\frac{\abs{a+b}}{1+\abs{a+b}}.
\end{align*}

hbghlyj

How to solve Rudin Chapter 2 problem 11?
Showing ρ(x,y)=d(x,y)/(1+d(x,y)) is a metric
Showing ρ(x,y)=d(x,y)/(1+d(x,y)) is a distance
Proving d_1=d(x,y)/(1+d(x,y)) is a metric equivalent with d, a given metric

hbghlyj

设$f(x)=\frac{\abs x}{1+\abs x}$，
相当于证明$f(x)=\frac{\abs x}{1+\abs x}$是subadditive（次加的），即$f(a)+f(b)\geqslant f(a+b),\forall a,b\inR$.
次加性和凹凸性有关：If a function $f$ is concave, and $f(0) ≥ 0$, then $f$ is subadditive on $[0,\infty)$.
$f(x)=\frac{\abs x}{1+\abs x}$连续但导数不连续：

Czhang271828

Approach Zero: A math-aware search engine.

hbghlyj

$f:\Bbb R\to\Bbb R;x\mapsto\frac{\abs x}{1+\abs x}$不是凸的，但是拟凸的。

取两个点$(-0.2,f(-0.2)),(2,f(2))$画一条线段，有一段在$f(x)$函数图象下方，故$f$不是凸的：
来自Wikipedia的配图：A quasiconvex function that is not convex 和上面$f(x)$的图象相近！