Đáp án: $P\ge \dfrac32$
Giải thích các bước giải:
Ta có :
$P=\dfrac{ab}{c^2\left(a+b\right)}+\dfrac{ac}{b^2\left(a+c\right)}+\dfrac{bc}{a^2\left(b+c\right)}$
$\to P=\dfrac{\left(\dfrac1c\right)^2}{\dfrac1a+\dfrac1b}+\dfrac{\left(\dfrac1b\right)^2}{\dfrac1a+\dfrac1c}+\dfrac{\left(\dfrac1a\right)^2}{\dfrac1c+\dfrac1b}$
$\to P\ge \dfrac{\left(\dfrac1c+\dfrac1b+\dfrac1a\right)^2}{\dfrac1a+\dfrac1b+\dfrac1a+\dfrac1c+\dfrac1c+\dfrac1b}$
$\to P\ge\dfrac{\left(\dfrac1a+\dfrac1b+\dfrac1c\right)^2}{2\left(\dfrac1a+\dfrac1b+\dfrac1c\right)}$
$\to P\ge \dfrac12\left(\dfrac1a+\dfrac1b+\dfrac1c\right)$
$\to P\ge \dfrac12.\dfrac9{a+b+c}$
$\to P\ge \dfrac32$
Dấu = xảy ra khi $a=b=c=1$
