[tex]\left ( \dfrac{x+1}{y} \right )^3+8+8\geq \dfrac{12(x+1)}{y}[/tex] [tex]\Rightarrow P \geq \dfrac{48(x+1)}{y}+\dfrac{48(y+1)}{x}+\sqrt{x^2+y^2}-128[/tex]
[tex]P \geq 48\left ( \dfrac{x}{y}+\dfrac{y}{x} \right )+48\left ( \dfrac{1}{x}+\dfrac{1}{y} \right )+\sqrt{x^2+y^2}-128 \geq 48\left ( \dfrac{1}{x}+\dfrac{1}{y} \right )+\sqrt{2xy}-32[/tex]
[tex]P \geq 48\left ( \dfrac{x+y}{xy} \right )+\sqrt{2xy}-32=48\left ( \dfrac{3-xy}{xy} \right )+\sqrt{2xy}-32=\dfrac{144}{xy}+\sqrt{2xy}-80[/tex]
Từ giả thiết: [tex]3=xy+x+y \geq xy+2\sqrt{xy}\Leftrightarrow (\sqrt{xy}-1)(\sqrt{xy}+3) \leq 0\Leftrightarrow \sqrt{xy}\leq 1\Leftrightarrow \sqrt{2xy} \leq \sqrt{2}[/tex]
Đặt [tex]\sqrt{2xy}=a\Rightarrow 0< a\leq \sqrt{2}[/tex]
[tex]P=\dfrac{288}{a^2}+a-80=\dfrac{\sqrt{2}}{a^2}+\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{288-\sqrt{2}}{a^2}-80 \geq 3\sqrt[3]{\frac{a^2}{2\sqrt{2}a^2}}+\dfrac{288-\sqrt{2}}{2}-80=64+\sqrt{2}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
m.n giúp em c15 với ạ, em cảm ơn