Cảm thấy bài của ''chị Anh'' có gì đó không ổn :D
#Fix
ĐK:\(\left\{{}\begin{matrix}ab+b+c\ne0\\ac+c+a\ne0\\bc+b+c\ne0\end{matrix}\right.\)
Áp dụng BĐT Cauchy-Schwarz dạng phân thức, ta có:
\(ab.\frac{1}{ab+a+b}\le ab.\frac{1}{9}\left(\frac{1}{ab}+\frac{1}{a}+\frac{1}{b}\right)\)\(=\frac{1}{9}+\frac{a}{9}+\frac{b}{9}\)
Tương tự: \(\frac{2ac}{ac+c+a}\le\frac{2}{9}+\frac{2a}{9}+\frac{2c}{9}\),\(\frac{3bc}{bc+b+c}\le\frac{3}{9}+\frac{3b}{9}+\frac{3c}{9}\)
Cộng vế theo vế, ta có;
\(\frac{ab}{ab+a+b}+\frac{2ac}{ac+c+a}+\frac{3bc}{bc+b+c}\le\frac{2}{3}+\frac{3a+4b+5c}{9}\)\(=\frac{2}{3}+\frac{12}{9}=2\)
\(''=''\Leftrightarrow a=b=c=1\)
ĐK: \(\left\{{}\begin{matrix}ab+a+b\ne0\\ac+a+c\ne0\\bc+b+c\ne0\end{matrix}\right.\)
Áp dụng BĐT Cô-si:
\(a+b\ge2ab\);\(a+c\ge2ac\);\(b+c\ge2bc\)
\(\Rightarrow A=\dfrac{ab}{ab+a+b}+\dfrac{3bc}{bc+b+c}+\dfrac{2ca}{ca+c+a}\)\(\le\dfrac{ab}{3ab}+\dfrac{2ac}{3ac}+\dfrac{3bc}{3bc}\)\(=\dfrac{1}{3}+\dfrac{2}{3}+1=2\)
Vậy Amax=2\(\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)\(\Rightarrow a=b=c\)và \(a,b,c\ne0\)
Thay vào 3a+4b+5c=12, ta có:
12a=12\(\Leftrightarrow a=b=c=1\)
