Giải thích các bước giải:
Ta có :
$P=\dfrac{\dfrac{1}{a^2}}{a(b+c)}+\dfrac{\dfrac{1}{b^2}}{b(c+a)}+\dfrac{\dfrac{1}{c^2}}{c(a+b)}$
$\rightarrow P=\dfrac{\dfrac{1}{a^2}}{a(b+c):(abc)}+\dfrac{\dfrac{1}{b^2}}{b(c+a):(abc)}+\dfrac{\dfrac{1}{c^2}}{c(a+b):(abc)}$
$\rightarrow P=\dfrac{(\dfrac{1}{a})^2}{\dfrac{1}{b}+\dfrac{1}{c}}+\dfrac{(\dfrac{1}{b})^2}{\dfrac{1}{c}+\dfrac{1}{a}}+\dfrac{(\dfrac{1}{c})^2}{\dfrac{1}{a}+\dfrac{1}{b}}$
$\rightarrow P\ge\dfrac{(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^2}{\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}}$
$\rightarrow P\ge\dfrac{(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})^2}{2(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})}$
$\rightarrow P\ge\dfrac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}{2}$
$\rightarrow P\ge\dfrac{3\sqrt[3]{\dfrac{1}{a}.\dfrac{1}{b}.\dfrac{1}{c}}}{2}$
$\rightarrow P\ge\dfrac{3}{2}$
Dấu = xảy ra khi $a=b=c=1$
