bài 5
[TEX]1=\sum \frac{1}{a^2}\geq \frac{1}{3}(\sum \frac{1}{a})^2=> \sum \frac{1}{a}\leq \sqrt3[/TEX]
[TEX]5a^2+2ab+2b^2=(2a+b)^2+(a-b)^2\geq (2a+b)^2[/TEX]
[TEX]=>\frac{1}{\sqrt{5a^2+2ab+2b^2}}\leq \frac{1}{2a+b}\leq \frac{1}{9}(\sum \frac{1}{a}) [/TEX]
[TEX]=> A\leq \frac{1}{3}(\sum \frac{1}{a})\leq \frac{\sqrt{3}}{3} [/TEX]