由三角不等式
[
|a-b|\le |a|+|b|=u+v,
]
所以
[
\left|\frac a{a-b}\right|\ge \frac{u}{u+v},
]
同理
[
\left|\frac b{b-c}\right|\ge \frac{v}{v+w},\qquad
\left|\frac c{c-a}\right|\ge \frac{w}{w+u}.
]
因此只需证
[
\frac{u}{u+v}+\frac{v}{v+w}+\frac{w}{w+u}\ge 1.
]
对它用柯西（Titu）即可：
[
\sum_{\text{cyc}}\frac{u}{u+v}
=\sum_{\text{cyc}}\frac{u^2}{u(u+v)}
\ge
\frac{(u+v+w)^2}{u(u+v)+v(v+w)+w(w+u)}.
]
而分母
[
u(u+v)+v(v+w)+w(w+u)
= u^2+v^2+w^2+uv+vw+wu
\le (u+v+w)^2.
]
所以
[
\sum_{\text{cyc}}\frac{u}{u+v}\ge 1.
]
于是
[
\sum_{\text{cyc}}\left|\frac a{a-b}\right|\ge 1.
]
[
|a-b|\le |a|+|b|=u+v,
]
所以
[
\left|\frac a{a-b}\right|\ge \frac{u}{u+v},
]
同理
[
\left|\frac b{b-c}\right|\ge \frac{v}{v+w},\qquad
\left|\frac c{c-a}\right|\ge \frac{w}{w+u}.
]
因此只需证
[
\frac{u}{u+v}+\frac{v}{v+w}+\frac{w}{w+u}\ge 1.
]
对它用柯西（Titu）即可：
[
\sum_{\text{cyc}}\frac{u}{u+v}
=\sum_{\text{cyc}}\frac{u^2}{u(u+v)}
\ge
\frac{(u+v+w)^2}{u(u+v)+v(v+w)+w(w+u)}.
]
而分母
[
u(u+v)+v(v+w)+w(w+u)
= u^2+v^2+w^2+uv+vw+wu
\le (u+v+w)^2.
]
所以
[
\sum_{\text{cyc}}\frac{u}{u+v}\ge 1.
]
于是
[
\sum_{\text{cyc}}\left|\frac a{a-b}\right|\ge 1.
]
