Đặt a + b - c = x; -a + b + c = y; a - b + c = z
=> \(\left\{{}\begin{matrix}x+y=2b\Rightarrow b=\dfrac{x+y}{2}\\y+z=2c\Rightarrow c=\dfrac{y+z}{2}\\z+x=2a\Rightarrow a=\dfrac{x+z}{2}\\x+y+z=a+b+c\end{matrix}\right.\)
Xét \(\dfrac{ab}{a+b-c}=\dfrac{\dfrac{x+z}{2}.\dfrac{x+y}{2}}{x}=\dfrac{\left(x+y\right)\left(x+z\right)}{4x}\)
Tương tự => \(\left\{{}\begin{matrix}\dfrac{bc}{-a+b+c}=\dfrac{\left(x+y\right)\left(y+z\right)}{4y}\\\dfrac{ac}{a-b+c}=\dfrac{\left(x+z\right)\left(y+z\right)}{4z}\end{matrix}\right.\)
\(VT=\dfrac{1}{4}\left[\dfrac{\left(x+y\right)\left(x+z\right)}{x}+\dfrac{\left(x+y\right)\left(y+z\right)}{y}+\dfrac{\left(x+z\right)\left(y+z\right)}{z}\right]\)
= \(\dfrac{1}{4}\left[\dfrac{x\left(x+y+z\right)+yz}{x}+\dfrac{y\left(x+y+z\right)+xz}{y}+\dfrac{z\left(x+y+z\right)+xy}{z}\right]\)
= \(\dfrac{1}{4}\left(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}+3x+3y+3z\right)\) (1)
Xét \(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}=\dfrac{1}{2}\left(2.\dfrac{yz}{x}+2.\dfrac{xz}{y}+2.\dfrac{xy}{z}\right)\)
= \(\dfrac{1}{2}\left[\left(\dfrac{yz}{x}+\dfrac{xz}{y}\right)+\left(\dfrac{xz}{y}+\dfrac{xy}{z}\right)+\left(\dfrac{xy}{z}+\dfrac{yz}{x}\right)\right]\)
Mà \(\dfrac{yz}{x}+\dfrac{xz}{y}\ge2\sqrt{\dfrac{yz}{x}.\dfrac{xz}{y}}=2z\)
\(\dfrac{xz}{y}+\dfrac{xy}{z}\ge2\sqrt{\dfrac{xz}{y}.\dfrac{xy}{z}}=2x\)
\(\dfrac{xy}{z}+\dfrac{yz}{x}\ge2\sqrt{\dfrac{xy}{z}.\dfrac{yz}{x}}=2y\)
=> \(\dfrac{yz}{x}+\dfrac{xz}{y}+\dfrac{xy}{z}\ge x+y+z\left(2\right)\)
(1)(2) => \(VT\ge a+b+c\)
Dấu "=" xảy ra <=> a = b = c
