Câu 3)
[tex]\sqrt{\frac{1}{a^2-ab+3b^2+1}}=\sqrt{\frac{1}{(a-b)^2+ab+(b^2+1)+b^2}}\leq \sqrt{\frac{1}{ab+2b+b^2}}=\sqrt{\frac{4}{4b(a+2+b)}}\leq \frac{1}{4b}+\frac{1}{a+2+b}\leq \frac{1}{4b}+\frac{1}{16}\left ( \frac{1}{a}+1+1+\frac{1}{b} \right )=\frac{5}{16b}+\frac{1}{16a}+\frac{1}{8}[/tex]
tương tự ta được : [tex]P \leq \frac{5}{16}\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )+\frac{1}{16}\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )+\frac{3}{8}\leq \frac{15}{16}+\frac{3}{16}+\frac{3}{8}=\frac{3}{2}[/tex]
dấu "=" khi a=b=c=1
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
