Cevian AA' bisects ∠BAC and therefore
|A'B| |A'C|
------- = -------
|BA| |CA|
or
|A'B| |CA'|
------- = -------
c b
Since |CA'| + |A'B| = |CB| = a,
b
--- |A'B| + |A'B| = a
c
or
ac
|A'B| = -----
b+c
Cevian BI bisects ∠ABA' and therefore
|IA| |IA'|
------ = -------
|AB| |A'B|
or
|AI| |IA'|
------ = ----------
c ac/(b+c)
Since |AI| + |IA'| = |AA'|,
a
|AI| + ----- |AI| = |AA'|
b+c
or
|AI| b+c
------- = -------
|AA'| a+b+c
Similarly,
|BI| c+a
------- = -------
|BB'| a+b+c
and
|CI| a+b
------- = -------
|CC'| a+b+c
Therefore,
|AI| |BI| |CI|
------- + ------- + -------
|AA'| |BB'| |CC'|
b+c c+a a+b
= ------- + ------- + -------
a+b+c a+b+c a+b+c
= 2
QED
See Harry's post for an alternate solution.
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