$ab\leq \frac{a^2+b^2}{2}=1$Áp dụng bdt Cauch-Schwarz$\Leftrightarrow \sum_{cyc}^{} \frac{a^4}{2016a^2+2017ab}\geq \frac{(a^2+b^2)^2}{2016(a^2+b^2)+4034ab}\geq \frac{2}{4033}$"=" khi $a=b=1$Đúng click "V" chấp nhận dùm Jin
$ab\leq \frac{a^2+b^2}{2}=1$Áp dụng bdt Cauch-Schwarz$\Leftrightarrow \sum_{cyc}^{} \frac{a^4}{2016a^2+2017ab}\geq \frac{(a^2+b^2)^2}{2016(a^2+b^2)+4034ab}\geq \frac{2}{4033}$"=" khi $a=b=1$
$ab\leq \frac{a^2+b^2}{2}=1$Áp dụng bdt Cauch-Schwarz$\Leftrightarrow \sum_{cyc}^{} \frac{a^4}{2016a^2+2017ab}\geq \frac{(a^2+b^2)^2}{2016(a^2+b^2)+4034ab}\geq \frac{2}{4033}$"=" khi $a=b=1$
Đúng click "V" chấp nhận dùm Jin