a, [tex]1+\sum \frac{1}{a}\geq 6(\sum \frac{1}{a^2})\geq 2(\sum \frac{1}{a})^2\Rightarrow 1\geq \sum \frac{1}{a}\geq \frac{-1}{2}\Rightarrow 1\geq \sum \frac{1}{a}[/tex]
Có [tex]\sum \frac{1}{10a+b+c}=\frac{1}{144}.\sum \frac{12^2}{10a+b+c}\leq \frac{1}{144}(\sum (\frac{10}{a}+\frac{1}{b}+\frac{1}{c}))=\frac{1}{144}.\sum \frac{12}{a}\leq \frac{1}{12}[/tex]
b,Có [tex]abc=a+b+c\Rightarrow \sum \frac{1}{ab}=1[/tex]
Đặt [tex]\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z[/tex] [tex]\Rightarrow \sum xy=1[/tex]
[tex]\bullet C/m\frac{3\sqrt{3}}{4}\leq \sum \frac{bc}{a(1+bc)}=\sum \frac{\frac{1}{a}}{\frac{1}{bc}+1}=\sum \frac{x}{yz+1}=A[/tex]
Có [tex]A=\sum \frac{x}{yz+1}=\sum \frac{x^2}{xyz+x}\geq ^{C-S}\frac{(x+y+z)^2}{3xyz+x+y+z}\geq ^{AM-GM}\frac{(x+y+z)^2}{\frac{(xy+yz+zx)^2}{x+y+z}+x+y+z}[/tex]
[tex]=\frac{(x+y+z)^3}{1+(x+y+z)^2}\geq \frac{3\sqrt{3}}{4}[/tex]
Hiển nhiên đúng do biến đổi tương đương
[tex]\bullet C/m:\sum \frac{bc}{a(1+bc)}\leq \frac{a+b+c}{4}[/tex]
Có [tex]\sum \frac{bc}{a(1+bc)}=\sum \frac{bc}{a+abc}=\sum \frac{bc}{a+a+b+c}\leq ^{AM-GM} \frac{1}{4}\sum (\frac{bc}{a+b}+\frac{bc}{a+c})=\frac{a+b+c}{4}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
\
