Đề 10 Đề thi thử vào 10 Vĩnh Bảo, Hải Phòng

28 Tháng ba 2019
343
953
71
14
Thái Bình
Lag..............Reconnect......Loading
Cho mình hỏi bài 4c và 5c làm thế nào vậy mọi người?
4c. Gợi ý: CM nó = 4R^2
5b.
[tex]\frac{1}{b+2c}+\frac{1}{b+2a}\geq \frac{4}{2a+2b+2c}=\frac{2}{a+b+c}\\\rightarrow \frac{a}{b+2c}+\frac{a}{b+2a}\geq \frac{2a}{a+b+c}(1)\\TT:\frac{b}{c+2a}+\frac{b}{c+2b}\geq \frac{2b}{a+b+c}(2)\\\frac{c}{a+2c}+\frac{c}{a+2b}\geq \frac{2c}{a+b+c}(3)\\(1)+(2)+(3):\\\rightarrow \frac{a}{b+2c}+\frac{a}{b+2a}+\frac{b}{c+2a}+\frac{b}{c+2b}+\frac{c}{a+2c}+\frac{c}{a+2b}\geq 2\\\rightarrow \frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\geq 2-\frac{a}{b+2a}-\frac{b}{c+2b}-\frac{c}{a+2c}\\\rightarrow 2(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b})\geq 4-\frac{2a}{b+2a}-\frac{2b}{c+2b}-\frac{2c}{a+2c}=1+1-\frac{2a}{b+2a}+1-\frac{2b}{c+2b}+1-\frac{2c}{a+2c}=1+\frac{b}{b+2a}+\frac{c}{c+2b}+\frac{a}{a+2c}(dpcm)[/tex]
 
Top Bottom