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
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
$$P= \frac{2(ab+bc+ca)}{2(2a+b+c)+abc}+\frac{8}{2a(b+c)+bc+4}-\frac{b+c+4}{\sqrt{bc}+1}$$
$$P+ \frac{7}{6}= \frac{2(ab+bc+ca)}{2(2a+b+c)+abc}- 1+\frac{8}{2a(b+c)+bc+4}-\frac{b+c+4}{\sqrt{bc}+1}+ \frac{13}{6}$$
$$\leq -\frac{a\left ( b- 2 \right )\left ( c- 2 \right )- 2\left (bc- b- c \right )}{a\left ( bc+ 4 \right )+ 2\left ( b+ c \right )}+ \frac{8}{2\left ( b+ c \right )+ bc+ 4}- \frac{b+ c+ 4}{\sqrt{bc+ 1}}+ \frac{13}{6}$$
$$b, c \in \left [ 1,2 \right ]\Rightarrow bc\leq b+ c$$
$$\Rightarrow \frac{8}{2\left ( b+ c \right )+ bc+ 4}- \frac{b+ c+ 4}{\sqrt{bc+ 1}}+ \frac{13}{6}\leq 0$$
$$\Rightarrow P\leq -\frac{7}{6}$$
$$P= \frac{2(ab+bc+ca)}{2(2a+b+c)+abc}+\frac{8}{2a(b+c)+bc+4}-\frac{b+c+4}{\sqrt{bc}+1}$$
$$P+ \frac{7}{6}= \frac{2(ab+bc+ca)}{2(2a+b+c)+abc}- 1+\frac{8}{2a(b+c)+bc+4}-\frac{b+c+4}{\sqrt{bc}+1}+ \frac{13}{6}$$
$$\leq -\frac{a\left ( b- 2 \right )\left ( c- 2 \right )- 2\left (bc- b- c \right )}{a\left ( bc+ 4 \right )+ 2\left ( b+ c \right )}+ \frac{8}{2\left ( b+ c \right )+ bc+ 4}- \frac{b+ c+ 4}{\sqrt{bc+ 1}}+ \frac{13}{6}$$
$$b, c \in \left [ 1,2 \right ]\Rightarrow bc\leq b+ c$$
$$\Rightarrow \frac{8}{2\left ( b+ c \right )+ bc+ 4}- \frac{b+ c+ 4}{\sqrt{bc+ 1}}+ \frac{13}{6}\leq 0$$
$$\Rightarrow P\leq -\frac{7}{6}$$
Mình thấy sai sai ...
Cho mình cáo lỗi:
Chỗ $\sqrt{bc+ 1}$ sửa lại $\sqrt{bc}+ 1$
Cho mình cáo lỗi:
Chỗ $\sqrt{bc+ 1}$ sửa lại $\sqrt{bc}+ 1$
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