Giải thích các bước giải:
Ta có :
$ab+bc+ca=abc\to \dfrac{1}{a}+\dfrac{1}b+\dfrac{1}c=1$
Đặt
$A=\dfrac{a}{bc(a+1)}+\dfrac{b}{ca(b+1)}+\dfrac{c}{ab(c+1)}$
$\to A=\dfrac{\dfrac{1}b.\dfrac{1}c}{1+\dfrac{1}a}+\dfrac{\dfrac{1}c.\dfrac{1}a}{1+\dfrac{1}b}+\dfrac{\dfrac{1}b.\dfrac{1}a}{1+\dfrac{1}c}$
Đặt $x= \dfrac{1}{a},y=\dfrac{1}b,z=\dfrac{1}c$
$\to x+y+z=1$
$A=\dfrac{xy}{z+1}+\dfrac{yz}{x+1}+\dfrac{zx}{y+1}$
Ta có :
$\dfrac{xy}{z+1}=\dfrac{xy}{(z+x)+(z+y)}\le \dfrac{1}{4}(\dfrac{xy}{x+z}+\dfrac{xy}{y+z})$
Tương tự ta chứng minh được
$\dfrac{yz}{x+1}\le \dfrac{yz}{x+y}+\dfrac{yz}{x+z}$
$\dfrac{zx}{y+1}\le \dfrac{zx}{x+y}+\dfrac{zx}{y+z}$
Cộng vế với vế
$\to A\le \dfrac{1}{4}(x+y+z)=\dfrac{1}4$
$\to$đpcm
