ta co $a^2-(b-c)^2=(a+b-c)(a-b+c)$ a+b-c=x>0
$b^2-(a-c)^2=(a+b-c)(-a+b+c)$ Dat a-b+c=y>0
$c^2-(a-b)^2=(a-b+c)(-a+b+c)$ -a+b+c+z>0
bat dang thuc ban dau tuong duong
$\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\leq$$\sqrt{(\frac{x+y}2)(\frac{x+z}2)}+\sqrt{(\frac{x+z}2)(\frac{y+z}2)}$$+\sqrt{(\frac{x+y}2)(\frac{x+z}2)}$
ma $\sqrt{(x+z)(x+y)}\geq\sqrt{(\sqrt{xy}+\sqrt{xz})^2}=\sqrt{xy}+\sqrt{xz}$
$\sqrt{(y+z)(x+y)}\geq\sqrt{xy}+\sqrt{yz}$
$\sqrt{(x+z)(y+z)}\geq\sqrt{xz}+\sqrt{yz}$
cong ba BDT tren ta co dieu phai CM