Đáp án: $P\le \dfrac14$
Giải thích các bước giải:
Ta có:
$P=\dfrac{ab}{c+1}+\dfrac{bc}{a+1}+\dfrac{ac}{b+1}$
$\to P=\dfrac{ab}{c+a+b+c}+\dfrac{bc}{a+a+b+c}+\dfrac{ac}{b+a+b+c}$
$\to P=\dfrac{ab}{(c+a)+(b+c)}+\dfrac{bc}{(a+b)+(c+a)}+\dfrac{ac}{(a+b)+(b+c)}$
$\to P=ab\cdot\dfrac{1}{(c+a)+(b+c)}+bc\cdot\dfrac{1}{(a+b)+(c+a)}+ca\cdot\dfrac{1}{(a+b)+(b+c)}$
$\to P\le ab\cdot \dfrac14(\dfrac1{c+a}+\dfrac1{b+c})+bc\cdot \dfrac14(\dfrac1{a+b}+\dfrac1{c+a})+ca\cdot \dfrac14(\dfrac1{a+b}+\dfrac1{b+c})$
$\to P\le \dfrac14(\dfrac{ab}{c+a}+\dfrac{ab}{b+c}+\dfrac{bc}{a+b}+\dfrac{bc}{c+a}+\dfrac{ca}{a+b}+\dfrac{ca}{b+c})$
$\to P\le \dfrac14((\dfrac{ab}{c+a}+\dfrac{bc}{c+a})+(\dfrac{ab}{b+c}+\dfrac{ca}{b+c})+(\dfrac{bc}{a+b}+\dfrac{ca}{a+b})$
$\to P\le \dfrac14(b+a+c)$
$\to P\le \dfrac14$
Dấu = xảy ra khi $a=b=c=\dfrac12$
