Đề bài sai .
b+c = x
a+c =y
b+a =z
\(a=\frac{y+z-x}{2};b=\frac{x+z-y}{2};c=\frac{x+y-z}{2}\) thay vào VT rồi áp dụng cosi => VP
làm tiếp
\(VT=\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}=\frac{1}{2}\left[\left(\frac{y}{x}+\frac{x}{y}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)+\left(\frac{z}{y}+\frac{y}{z}\right)-3\right]\ge\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)
=> dpcm