Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{2a+b}{c}=\frac{2b+c}{a}=\frac{2c+a}{b}=\frac{2a+b+2b+c+2c+a}{a+b+c}=\frac{3b+3c+3a}{a+b+c}=3\)

=>2a+b=3c; 2b+c=3a; 2c+a=3b

\(\left(\frac{2a+b}{c}\right)+\left(\frac{a}{2b+c}\right)+\left(\frac{3b}{2c+a}\right)\)

\(=\frac{3c}{c}+\frac{a}{3a}+\frac{3b}{3b}=3+\frac13+1=4+\frac13=\frac{13}{3}\)

Theo tính chất dãy tỉ số bằng nhau ta có:

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

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

Vậy \(\frac{2a+b}{c}=\frac{a}{2b+c}=\frac{3b}{2c+a}=1\)

vậy nếu a+b+c = 0 thì sao ?

Áp dụng t/c dãy tỉ số bằng nhau:

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

\(\Rightarrow\hept{\begin{cases}2a+b=3c\\2b+c=3a\\3c+a=3b\end{cases}}\)

\(\Rightarrow BT=\frac{3c}{c}+\frac{a}{3a}+\frac{3b}{b}=6+\frac{1}{3}=\frac{19}{3}\)

Vậy nếu a+b+c = 0 thì sao ?

\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Leftrightarrow\left\{{}\begin{matrix}2b+c-a=2a\\2c-b+a=2b\\2a+b-c=2c\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3a-2b=c\\3b-2c=a\\3c-2a=b\end{matrix}\right.\text{ và }\left\{{}\begin{matrix}3a-c=2b\\3b-a=2c\\3c-b=2a\end{matrix}\right.\\ \Leftrightarrow P=\dfrac{a\cdot b\cdot c}{2a\cdot2b\cdot3c}=\dfrac{1}{8}\)

\(\dfrac{a^2b^2}{2a^2+b^2+3a^2b^2}=\dfrac{a^2b^2}{\left(a^2+b^2\right)+\left(a^2+a^2b^2\right)+2a^2b^2}\le\dfrac{a^2b^2}{2ab+2a^2b+2a^2b^2}=\dfrac{ab}{2\left(1+a+ab\right)}\)

Tương tự và cộng lại;

\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{bc}{1+b+bc}+\dfrac{ca}{1+c+ca}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{abc}{a+ab+abc}+\dfrac{ab.ca}{ab+abc+ab.ca}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{1}{a+ab+1}+\dfrac{a}{ab+1+a}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)