[tex]5a^2+2ab+2b^2=(a^2-2ab+b^2)+(4a^2+4ab+b^2)=(a-b)^2+(2a+b)^2\geq (2a+b)^2\Rightarrow \frac{1}{\sqrt{5a^2+2ab+2b^2}}\leq \frac{1}{2a+b}=\frac{1}{a+a+b}\leq \frac{1}{9}(\frac{1}{a}+\frac{1}{a}+\frac{1}{b})[/tex]
Tương tự ta có: [tex]P \leq \frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})[/tex]
Mà [tex]1=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq \frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})^2\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq \sqrt{3}\Rightarrow P\leq \frac{\sqrt{3}}{3}[/tex]
Dấu "=" xảy ra tại [tex]a=b=c=\sqrt{3}[/tex]