Chủ đề này có 5 trả lời

[hoicmvsao]

Cho a,b,c $\in [1;2]$ ($1\leq a,b,c\leq 2$) , a,b,c>0

Tìm Max P =$\frac{2(ab+bc+ca)}{2(2a+b+c)+abc}+\frac{8}{2a(b+c)+bc+4}-\frac{b+c+4}{\sqrt{bc}+1}$

Bài viết đã được chỉnh sửa nội dung bởi hoicmvsao: 21-02-2018 - 19:19

[DOTOANNANG]

$$P= \frac{2(ab+bc+ca)}{2(2a+b+c)+abc}+\frac{8}{2a(b+c)+bc+4}-\frac{b+c+4}{\sqrt{bc}+1}$$

$$P+ \frac{7}{6}= \frac{2(ab+bc+ca)}{2(2a+b+c)+abc}- 1+\frac{8}{2a(b+c)+bc+4}-\frac{b+c+4}{\sqrt{bc}+1}+ \frac{13}{6}$$

$$\leq -\frac{a\left ( b- 2 \right )\left ( c- 2 \right )- 2\left (bc- b- c \right )}{a\left ( bc+ 4 \right )+ 2\left ( b+ c \right )}+ \frac{8}{2\left ( b+ c \right )+ bc+ 4}- \frac{b+ c+ 4}{\sqrt{bc+ 1}}+ \frac{13}{6}$$

$$b, c \in \left [ 1,2 \right ]\Rightarrow bc\leq b+ c$$

$$\Rightarrow \frac{8}{2\left ( b+ c \right )+ bc+ 4}- \frac{b+ c+ 4}{\sqrt{bc+ 1}}+ \frac{13}{6}\leq 0$$

$$\Rightarrow P\leq -\frac{7}{6}$$

 

[hoicmvsao]

$$P= \frac{2(ab+bc+ca)}{2(2a+b+c)+abc}+\frac{8}{2a(b+c)+bc+4}-\frac{b+c+4}{\sqrt{bc}+1}$$

$$P+ \frac{7}{6}= \frac{2(ab+bc+ca)}{2(2a+b+c)+abc}- 1+\frac{8}{2a(b+c)+bc+4}-\frac{b+c+4}{\sqrt{bc}+1}+ \frac{13}{6}$$

$$\leq -\frac{a\left ( b- 2 \right )\left ( c- 2 \right )- 2\left (bc- b- c \right )}{a\left ( bc+ 4 \right )+ 2\left ( b+ c \right )}+ \frac{8}{2\left ( b+ c \right )+ bc+ 4}- \frac{b+ c+ 4}{\sqrt{bc+ 1}}+ \frac{13}{6}$$

$$b, c \in \left [ 1,2 \right ]\Rightarrow bc\leq b+ c$$

$$\Rightarrow \frac{8}{2\left ( b+ c \right )+ bc+ 4}- \frac{b+ c+ 4}{\sqrt{bc+ 1}}+ \frac{13}{6}\leq 0$$

$$\Rightarrow P\leq -\frac{7}{6}$$

Mình thấy sai sai ...

[DOTOANNANG]

Cho mình cáo lỗi:

Chỗ $\sqrt{bc+ 1}$ sửa lại $\sqrt{bc}+ 1$

[hoicmvsao]

Cho mình cáo lỗi:

Chỗ $\sqrt{bc+ 1}$ sửa lại $\sqrt{bc}+ 1$

Dấu bằng xảy ra khi nào vậy bạn?

[DOTOANNANG]

$$\boldsymbol{a= 1, b= c= 2}$$