https://h.vn/hoi-dap/question/702421.html
https://h.vn/hoi-dap/question/702421.html
https://h.vn/hoi-dap/question/702421.html
Ta có:
\(\sqrt{12a+\left(b-c\right)^2}=\sqrt{4a\left(a+b+c\right)+\left(b-c\right)^2}\)
\(=\sqrt{4a^2+4ab+4ac+b^2-2bc+c^2}\)
\(=\sqrt{\left(2a+b+c\right)^2-4bc}\)
\(\le\sqrt{\left(2a+b+c\right)^2}=2a+b+c\)
Khi đó \(K\le4\left(a+b+c\right)=12\)
Dấu "=" xảy ra tại \(a=0;b=0;c=3\) và các hoán vị.
A no thơ guây:v
\(12a+\left(b-c\right)^2=12a+\left(b+c\right)^2-4bc\)
\(=\left(a-3\right)^2+12a-4bc\)
\(=a^2-6a+9+12a-4bc\)
\(=a^2+6a+9-4bc\)
\(=\left(a+3\right)^2-4bc\le\left(a+3\right)^2\)
\(\Rightarrow\sqrt{12a+\left(b-c\right)^2}\le a+3\)
Cộng lại ta có:
\(K\le a+b+c+9=12\)
Dấu "=" xảy ra.....
