theo cái t nói ấy
[tex]\sqrt{3a^2+8b^2+14ab}\leq 2a+3b[/tex] (cm bằng cách bình phương)
suy ra
[tex]\sum \frac{a^2}{\sqrt{3a^2+8b^2+14ab}}\geq \sum \frac{a^2}{2a+3b}\\\geq \frac{(a+b+c)^2}{5(a+b+c)}=\frac{a+b+c}{5}[/tex]
[tex]\sqrt{3a^2+8b^2+14ab}\leq 2a+3b[/tex] (cm bằng cách bình phương)
suy ra
[tex]\sum \frac{a^2}{\sqrt{3a^2+8b^2+14ab}}\geq \sum \frac{a^2}{2a+3b}\\\geq \frac{(a+b+c)^2}{5(a+b+c)}=\frac{a+b+c}{5}[/tex]
