Đáp án: $P\le \dfrac14$
Giải thích các bước giải:
Ta có:
$ab+bc+ca=abc\to\dfrac1a+\dfrac1b+\dfrac1c=1$
Đặt $\dfrac1a=x,\dfrac1b=y,\dfrac1c=z\to x+y+z=1, (x,y,z>0)$
Ta có:
$P=\dfrac{a}{bc(a+1)}+\dfrac{b}{ac(b+1)}+\dfrac{c}{ab(c+1)}$
$\to P=\dfrac{\dfrac1b.\dfrac1c}{1+\dfrac1a}+\dfrac{\dfrac1a.\dfrac1c}{1+\dfrac1b}+\dfrac{\dfrac1a.\dfrac1b}{1+\dfrac1c}$
$\to P=\dfrac{yz}{1+x}+\dfrac{xz}{1+y}+\dfrac{xy}{1+z}$
$\to P=\dfrac{yz}{x+y+z+x}+\dfrac{xz}{x+y+z+y}+\dfrac{xy}{x+y+z+z}$
$\to P=yz\cdot \dfrac{1}{(y+x)+(z+x)}+xz\cdot \dfrac{1}{(x+y)+(z+y)}+xy\cdot \dfrac{1}{(x+z)+(y+z)}$
$\to P\le yz\cdot \dfrac14(\dfrac{1}{x+y}+\dfrac{1}{z+x})+xz\cdot \dfrac14(\dfrac{1}{x+y}+\dfrac{1}{y+z})+xy\cdot\dfrac14(\dfrac{1}{x+z}+\dfrac{1}{y+z})$
$\to P\le \dfrac14(\dfrac{yz}{x+y}+\dfrac{yz}{z+x}+\dfrac{xz}{x+y}+\dfrac{xz}{y+z}+\dfrac{xy}{x+z}+\dfrac{xy}{y+z})$
$\to P\le \dfrac14(\dfrac{yz+xz}{x+y}+\dfrac{yz+xy}{z+x}+\dfrac{xz+xy}{y+z})$
$\to P\le \dfrac14(z+y+x)$
$\to P\le \dfrac14$
Dấu = xảy ra khi $x=y=z=\dfrac13$
$\to a=b=c=3$
