Đặt A= \(\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\)
Đặt x = b + c - a, y = a + c - b, z =a + b -c
=>\(\left\{{}\begin{matrix}x+y=2c\\y+z=2a\\x+z=2b\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}c=\dfrac{x+y}{2}\\a=\dfrac{y+z}{2}\\b=\dfrac{x+z}{2}\end{matrix}\right.\)
\(\Leftrightarrow A=2\left(\dfrac{\dfrac{y+z}{2}}{x}+\dfrac{\dfrac{x+z}{2}}{y}+\dfrac{\dfrac{x+y}{2}}{z}\right)\)
\(\Leftrightarrow A=\dfrac{y+z}{x}+\dfrac{x+z}{y}+\dfrac{x+y}{z}\)
\(\Leftrightarrow A=\dfrac{y}{x}+\dfrac{x}{y}+\dfrac{z}{x}+\dfrac{x}{z}+\dfrac{z}{y}+\dfrac{y}{z}\)
Theo bất đẳng thức Cô -si luôn đúng với m, n \(\ge0\)
=> \(m+n\ge2\sqrt{m.n}\) . Dấu '=' xảy ra kh m = n
=> Ta có : \(\left\{{}\begin{matrix}\dfrac{y}{x}+\dfrac{x}{y}\ge2\sqrt{\dfrac{y}{x}.\dfrac{x}{y}}=2\left(1\right)\\\dfrac{z}{x}+\dfrac{x}{z}\ge2\sqrt{\dfrac{z}{x}.\dfrac{x}{z}}=2\\\dfrac{z}{y}+\dfrac{y}{z}\ge2\sqrt{\dfrac{z}{y}.\dfrac{y}{z}}=2\left(3\right)\end{matrix}\right.\left(2\right)\)
Cộng từng vế 3 bất đẳng thức (1) (2) (3) , ta được:
A \(\ge6\)
Vậy \(\dfrac{2a}{b+c-a}+\dfrac{2b}{a+c-b}+\dfrac{2c}{a+b-c}\ge6.\)Dấu '=' xảy ra khi a = b =c.
\(\dfrac{2a}{b+c-a}+\dfrac{2b}{c+a-b}+\dfrac{2c}{a+b-c}\)
\(=\dfrac{2a^2}{ab+ac-a^2}+\dfrac{2b^2}{ba+bc-b^2}+\dfrac{2c^2}{ca+cb-c^2}\)
\(\ge\dfrac{2\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)-a^2-b^2-c^2}\)
\(\ge\dfrac{2\left(a+b+c\right)^2}{\dfrac{\left(a+b+c\right)^2}{3}+a^2+b^2+c^2-a^2-b^2-c^2}=6\)
Dấu = xảy ra khi a = b = c
