\(M=\dfrac{ab}{bc+ca}+\dfrac{ac}{ab+bc}+\dfrac{bc}{ab+ca}\)
\(M=\dfrac{\left(ab\right)^2}{ab^2c+a^2bc}+\dfrac{\left(ca\right)^2}{a^2bc+abc^2}+\dfrac{\left(bc\right)^2}{ab^2c+abc^2}\)
\(M\ge\dfrac{\left(ab+bc+ca\right)^2}{2abc\left(a+b+c\right)}\ge\dfrac{3abc\left(a+b+c\right)}{2abc\left(a+b+c\right)}=\dfrac{3}{2}\)
\(M_{min}=\dfrac{3}{2}\) khi \(a=b=c=1\)
Ở bước đầu tiên nếu mình đặt ẩn phụ \(\left(ab;bc;ca\right)=\left(x;y;z\right)\) thì M là Netsbitt
Áp dụng BĐT Cói và BĐT BSC:
\(\dfrac{ab}{c^2\left(a+b\right)}+\dfrac{a+b}{4ab}\ge\dfrac{1}{c}\Rightarrow\dfrac{ab}{c^2\left(a+b\right)}\ge\dfrac{1}{c}-\dfrac{1}{4a}-\dfrac{1}{4b}\)
Tương tự: \(\dfrac{bc}{a^2\left(b+c\right)}\ge\dfrac{1}{a}-\dfrac{1}{4b}-\dfrac{1}{4c};\dfrac{ca}{b^2\left(c+a\right)}\ge\dfrac{1}{b}-\dfrac{1}{4c}-\dfrac{1}{4a}\)
\(\Rightarrow M\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{9}{2\left(a+b+c\right)}=\dfrac{3}{2}\)
\(minM=\dfrac{3}{2}\Leftrightarrow a=b=c=1\)