Ta có:[tex]P=\frac{6ab}{a+b}-a^2-b^2\leq \frac{6ab}{a+b}-2ab\leq \frac{6ab}{2\sqrt{ab}}-2ab=3\sqrt{ab}-2ab[/tex]
Lại có:[tex]1=3ab+a+b\geq 3ab+2\sqrt{ab}\Rightarrow 3ab+2\sqrt{ab}-1\leq 0\Rightarrow (3\sqrt{ab}-1)(\sqrt{ab}+1)\leq 0\Rightarrow \sqrt{ab}\leq \frac{1}{3}[/tex]
[tex]\Rightarrow P\leq 3\sqrt{ab}-2ab=-2(ab-\frac{2}{3}\sqrt{ab}+\frac{1}{9})+\frac{5}{3}\sqrt{ab}+\frac{2}{9}=\frac{5}{3}\sqrt{ab}-2(\sqrt{ab}-\frac{1}{3})^2+\frac{2}{9}\leq \frac{5}{3}.\frac{1}{3}+\frac{2}{9}=\frac{7}{9}[/tex]
Dấu "=" xảy ra tại [tex]x=y=\frac{1}{3}[/tex]
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