sửa đề : cho a,b,c là 3 cạnh tam giác : CM : ab/a+b-c + bc/-a+b+c + ac/a-b+c \(\ge\)a+b+c
vì a,b,c là 3 cạnh của 1 tam giác nên a + b - c > 0 ; -a +b + c > 0 ; a - b + c > 0
Đặt x = a + b - c ; y = -a + b + c ; z = a - b + c
Ta có : x + y + z = a + b + c ; a = \(\frac{y+z}{2}\); b = \(\frac{x+z}{2}\); c = \(\frac{x+y}{2}\)
\(\frac{ab}{a+b-c}+\frac{bc}{-a+b+c}+\frac{ac}{a-b+c}=\frac{\left(y+z\right).\left(x+z\right)}{4z}+\frac{\left(x+z\right).\left(x+y\right)}{4x}+\frac{\left(x+y\right).\left(y+z\right)}{4y}\)
\(=\frac{1}{4}.\left(\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}+3x+3y+3z\right)\)
\(=\frac{1}{4}.\left[3.\left(x+y+z\right)+\frac{1}{2}.\left(2\frac{xy}{z}+2\frac{yz}{x}+2\frac{xz}{y}\right)\right]\)
\(=\frac{1}{4}.\left[3.\left(x+y+z\right)+\frac{y}{2}.\left(\frac{x}{z}+\frac{z}{x}\right)+\frac{x}{2}.\left(\frac{y}{z}+\frac{z}{y}\right)+\frac{z}{2}.\left(\frac{x}{y}+\frac{y}{x}\right)\right]\)
\(\ge\frac{1}{4}.\left[3.\left(x+y+z\right)+x+y+z\right]=x+y+z\)
Mà x + y + z = a + b + c
\(\Rightarrow\)\(\frac{ab}{a+b-c}+\frac{bc}{-a+b+c}+\frac{ac}{a-b+c}\)\(\ge\)\(a+b+c\)
