[tex]P=9-3(a+b)+ab\\a^2+b^2=1\rightarrow (a+b)^2=1+2ab\\\rightarrow a+b=\sqrt{1+2ab}\\\rightarrow P=9-3\sqrt{1+2ab}+ab\\1+2ab+2\geq 2\sqrt{2(1+2ab)}\rightarrow \sqrt{1+2ab}\leq \frac{3+2ab}{2\sqrt{2}}\\\rightarrow P\geq 9-3.\frac{3+2ab}{2\sqrt{2}}+ab=9-\frac{9}{2\sqrt{2}}-\frac{3ab}{\sqrt{2}}+ab=9-\frac{9}{2\sqrt{2}}-\frac{(3-\sqrt{2})ab}{\sqrt{2}}\\a^2+b^2\geq 2ab\rightarrow ab\leq \frac{1}{2}\\\rightarrow P\geq 9-\frac{9}{2\sqrt{2}}-\frac{(3-\sqrt{2})\frac{1}{2}}{\sqrt{2}}=9-\frac{9}{2\sqrt{2}}-\frac{3}{2\sqrt{2}}+\frac{1}{2}=9-\frac{6}{\sqrt{2}}+\frac{1}{2}=(3-\frac{1}{\sqrt{2}})^2[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
