Cho $a,b,c>0$, thoả mãn:$a^{4}+b^{4}+c^{4}=3$. Tìm Max
$A=\frac{1}{4-ab}+\frac{1}{4-bc}+\frac{1}{4-ac}$
Ta có :
$2A= \sum \frac{2}{4-ab}=\sum (1-\frac{2-ab}{4-ab})= \sum (1-\frac{4-a^{2}b^{2}}{(4-ab)(2+ab)})= \sum (1-\frac{4-a^{2}b^{2}}{-(ab-1)^{2}+9})\leq\sum (1-\frac{4-a^{2}b^{2}}{9})= \frac{15}{9}+\sum \frac{a^{2}b^{2}}{9}\leq \frac{15}{9}+\frac{\sum a^{4}}{9}=2$
Vậy $MaxA=1$ , dấu "=" xảy ra khi $a=b=c=1$