Giải thích các bước giải:
Ta có : $(x+y)^2\ge 4xy\to\dfrac{x+y}{xy}\ge \dfrac{4}{x+y}\to\dfrac{1}{x}+\dfrac{1}{y}\ge \dfrac{4}{x+y}$
Ta có :
$A=\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ca}{b+1}$
$\to A=\dfrac{ab}{c+a+b+c}+\dfrac{bc}{a+a+b+c}+\dfrac{ca}{b+a+b+c}$
$\to A=\dfrac{ab}{(a+c)+(b+c)}+\dfrac{bc}{(b+a)+(c+a)}+\dfrac{ca}{(b+a)+(c+b)}$
$\to A=ab.\dfrac{1}{(a+c)+(b+c)}+bc.\dfrac{1}{(b+a)+(c+a)}+ca.\dfrac{1}{(b+a)+(c+b)}$
$\to A\le ab.\dfrac 14(\dfrac{1}{a+c}+\dfrac{1}{b+c})+bc.\dfrac{1}{4}(\dfrac{1}{b+a}+\dfrac{1}{c+a})+ca.\dfrac 14(\dfrac{1}{b+a}+\dfrac{1}{c+b})$
$\to A\le \dfrac 14(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{bc}{b+a}+\dfrac{bc}{c+a}+\dfrac{ca}{b+a}+\dfrac{ca}{c+b})$
$\to A\le \dfrac 14(\dfrac{ab+bc}{a+c}+\dfrac{ab+ca}{b+c}+\dfrac{bc+ca}{b+a})$
$\to A\le \dfrac 14(a+b+c)=\dfrac 14$
Dấu = xảy ra khi $a=b=c=\dfrac 13$
