Giải thích các bước giải:
Ta có:
$\dfrac{1}{a^2(b+c)}=\dfrac{1}{a(ab+ac)}=\dfrac1a\cdot \dfrac1{ab+ac}=\dfrac1a\cdot \dfrac1{\dfrac1c+\dfrac1b}=\dfrac{\dfrac1a}{\dfrac1b+\dfrac1c}$ vì $abc=1$
Tương tự có:
$\dfrac{1}{b^2(a+c)}=\dfrac{\dfrac1b}{\dfrac1c+\dfrac1a}$
$\dfrac1{c^2(a+b)}=\dfrac{\dfrac1c}{\dfrac1b+\dfrac1a}$
$\to $Bất đẳng thức cần chứng minh tương đương
$P=\dfrac{\dfrac1a}{\dfrac1b+\dfrac1c}+\dfrac{\dfrac1b}{\dfrac1c+\dfrac1a}+\dfrac{\dfrac1c}{\dfrac1b+\dfrac1a}$
Đặt $\dfrac1a=x, \dfrac1b=y, \dfrac1c=z$
$\to P=\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{y+x}$
$\to P=\dfrac{x^2}{xy+xz}+\dfrac{y^2}{zy+xy}+\dfrac{z^2}{yz+xz}$
$\to P\ge \dfrac{(x+y+z)^2}{xy+xz+zy+xy+yz+xz}$
$\to P\ge \dfrac{(x+y+z)^2}{2(xy+yz+zx)}$
$\to P\ge \dfrac{3(xy+yz+zx)}{2(xy+yz+zx)}$
$\to P\ge \dfrac32$
$\to đpcm$
