Question

[Nguyễn Trần Trúc Ly]

Hỡi các bác giỏi bất đẳng thức!!!!

Answer 1

[Nguyễn Việt Lâm]

\(abc=3b+6c\Leftrightarrow a=\frac{3}{c}+\frac{6}{b}\Rightarrow\frac{a}{3}=\frac{1}{c}+\frac{2}{b}\)

\(P=\frac{3}{b+c-a}+\frac{4}{c+a-b}+\frac{5}{a+b-c}\)

\(P=\frac{1}{b+c-a}+\frac{1}{c+a-b}+2\left(\frac{1}{b+c-a}+\frac{1}{a+b-c}\right)+3\left(\frac{1}{c+a-b}+\frac{1}{a+b-c}\right)\)

\(P\ge\frac{4}{b+c-a+c+a-b}+2.\frac{4}{b+c-a+a+b-c}+3.\frac{4}{c+a-b+a+b-c}\)

\(P\ge\frac{2}{c}+\frac{4}{b}+\frac{6}{a}=2\left(\frac{1}{c}+\frac{2}{b}\right)+\frac{6}{a}=\frac{2a}{3}+\frac{6}{a}\ge2\sqrt{\frac{2a}{3}.\frac{6}{a}}=4\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=3\)

Answer 2

[Nguyễn Việt Lâm]

Bài còn lại đơn giản hơn nhiều:

\(2ab+6bc+2ac=7abc\Leftrightarrow\frac{2}{c}+\frac{6}{a}+\frac{2}{b}=7\)

\(C=\frac{4ab}{a+2b}+\frac{9ac}{a+4c}+\frac{4ab}{b+c}=\frac{4}{\frac{1}{b}+\frac{2}{a}}+\frac{9}{\frac{1}{c}+\frac{4}{a}}+\frac{4}{\frac{1}{b}+\frac{1}{c}}\)

\(C=\frac{2^2}{\frac{1}{b}+\frac{2}{a}}+\frac{3^2}{\frac{1}{c}+\frac{4}{a}}+\frac{2^2}{\frac{1}{b}+\frac{1}{c}}\ge\frac{\left(2+3+2\right)^2}{\frac{1}{b}+\frac{2}{a}+\frac{1}{c}+\frac{4}{a}+\frac{1}{b}+\frac{1}{c}}\)

\(C\ge\frac{49}{\frac{2}{c}+\frac{6}{a}+\frac{2}{b}}=\frac{49}{7}=7\)

\(\Rightarrow C_{min}=7\) khi \(\left\{{}\begin{matrix}b=c=1\\a=2\end{matrix}\right.\)