Ta có :
$P = \dfrac{a}{a+1} + \dfrac{b}{b+1}+\dfrac{c}{c+1}$
$= 1-(\dfrac{a^2}{a+1}+\dfrac{b^2}{b+1}+\dfrac{c^2}{c+1}) $
$≤ 1 - \dfrac{(a+b+c)^2}{a+1+b+1+c+1}$
$= 1- \dfrac{1}{4} = \dfrac{3}{4}$
Dấu "=" xảy ra $⇔a=b=c=\dfrac{1}{3}$
Vậy : $P_{min} = \dfrac{3}{4}$ tại $a=b=c=\dfrac{1}{3}$