Từ $xy +y -3x+1 =0$ đặt a=$\frac{1}{X}$
Ta được $(a+1)(y+1)=4 <=>S+P=3=>P=3-S=>S\geq 2$
$P=8((\frac{a}{y+3})^{3}+(\frac{y}{a+3})^{3})-\sqrt{a^{2}+y^{2}}\geq 2(\frac{a}{y+3}+\frac{y}{a+3})^{3}-\sqrt{a^{2}+y^{2}}=2(\frac{S^{2}+3S-2P}{3S+P+9})^{3}-\frac{S}{\sqrt{2}}=2(\frac{S^{2}+3S-2(3-S)}{3S+(3-S) +9})^{3}-\frac{S}{\sqrt{2}}=2(\frac{S^{2}+5S-6}{2S+12})^{3}-\frac{S}{\sqrt{2}}=2(\frac{S-1}{2})^{2}-\frac{S}{\sqrt{2}}=\frac{(S-1)^{3}}{4}-\frac{S}{\sqrt{2}},S\geq 2$
f'(S)=$\frac{3}{4}(S-1)^{2}-\frac{1}{\sqrt{2}}>0,\forall S\geq 2=>min P=f(2)=\frac{1}{4}-\sqrt{2}$
dấu bằng xảy ra khi a=y=1 hay x=y=1
chúc các bạn học tốt!