Câu hỏi của vvvvvvvv

Question

cho a,b,c > 0 và a+b+c$\le3$

chứng minh rằng B=$\frac{1}{a^2+b^2+c^2}+\frac{2009}{ab+bc+ac}\ge670$

Nguyễn Việt Lâm · Accepted Answer

$B=\frac{1}{a^2+b^2+c^2}+\frac{4}{2ab+2bc+2ac}+\frac{2007}{ac+bc+ac}$

$B\ge\frac{\left(1+2\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}+\frac{2007}{\frac{\left(a+b+c\right)^2}{3}}$

$B\ge\frac{9}{\left(a+b+c\right)^2}+\frac{6021}{\left(a+b+c\right)^2}\ge\frac{9}{3^2}+\frac{6021}{3^2}=670$

Dấu "=" xảy ra khi $a=b=c=1$Câu trả lời: $B=\frac{1}{a^2+b^2+c^2}+\frac{4}{2ab+2bc+2ac}+\frac{2007}{ac+bc+ac}$

$B\ge\frac{\left(1+2\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}+\frac{2007}{\frac{\left(a+b+c\right)^2}{3}}$

$B\ge\frac{9}{\left(a+b+c\right)^2}+\frac{6021}{\left(a+b+c\right)^2}\ge\frac{9}{3^2}+\frac{6021}{3^2}=670$

Dấu "=" xảy ra khi $a=b=c=1$

Vu Ngoc Chau · Answer

ý 2 là sao vậy bạnCâu trả lời: ý 2 là sao vậy bạn