$1)$ Đặt $a = \dfrac{1}{x} ; b=\dfrac{1}{y} ; c = \dfrac{1}{z}$
$\Rightarrow x.y.z =1$
BĐT : $\Leftrightarrow \dfrac{x^2}{\frac{1}{y}+ \frac{1}{z}}+ \dfrac{y^2}{\frac{1}{z} +\frac{1}{x}} + \dfrac{z^2}{\frac{1}{x}+ \frac{1}{y}}$ `>=` $\dfrac{3}{2}$
Áp dụng $Nesbitt$ :
$\Leftrightarrow \dfrac{1}{a^2(b+c)} + \dfrac{1}{b^2(a+c)}+\dfrac{1}{c^2(a+b)} > \dfrac{3}{2}$
$2)$
Ta có : $(a+b+c). (\dfrac{a}{(b+c)^2} + \dfrac{b}{(c+a)^2} + \dfrac{c}{(a+b)^2}$ `>=` $(\dfrac{a}{b+c} + \dfrac{b} {a+c} + \dfrac {c} {a+b})^2$
Phân tích vế phải :
$\dfrac{a}{b+c} + \dfrac{b} {a+c} + \dfrac {c} {a+b} = \dfrac{a^2}{ab+ac} + \dfrac{b^2}{bc+ba}+ \dfrac {c^2}{ac+ bc}$ `>=` $\dfrac{(a+b+c)^2}{2(ab+ac+bc)}$ `>=` $\dfrac{3}{2}$
$\Rightarrow \text{đpcm}$. Điều kiện dấu '=' xảy ra $\Leftrightarrow a=b=c$
