đặt : \(x=b+c,y=c+a,z=a+b\) \(\left(a>0,b>0,c>0\right)\)
\(\Rightarrow2a=y-c+z-b,2b=x-c+z-a,2c=x-b+y-a\)
\(\Rightarrow2a=y+z-x,2b=x+z-y,2c=x+y-z\)
\(\Rightarrow a=\frac{y+z-x}{2}\left(1\right),b=\frac{x+z-y}{2}\left(2\right),c=\frac{x+y-z}{2}\left(3\right)\)
thay (1),(2),(3) vào biểu thức\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\left(a>0,b>0,c>0\right)\) ta có :
\(\frac{\frac{y+z-x}{2}}{x}+\frac{\frac{x+z-y}{2}}{y}+\frac{\frac{x+y-z}{2}}{z}\)
\(=\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\)
\(=\frac{1}{2}\frac{y+z}{x}-\frac{1}{2}+\frac{1}{2}\frac{x+z}{y}-\frac{1}{2}+\frac{1}{2}\frac{x+y}{z}-\frac{1}{2}\)
\(=\frac{1}{2}\frac{y+z}{x}+\frac{1}{2}\frac{x+z}{y}+\frac{1}{2}\frac{x+y}{z}-\frac{3}{2}\)
\(=\frac{1}{2}\left(\frac{y}{x}+\frac{z}{x}+\frac{x}{y}+\frac{z}{y}+\frac{x}{z}+\frac{y}{z}\right)-\frac{3}{2}\)
\(=\frac{1}{2}\left(\frac{y}{x}+\frac{x}{y}+\frac{z}{x}+\frac{x}{z}+\frac{z}{y}+\frac{y}{z}\right)-\frac{3}{2}\ge\frac{1}{2}\cdot6-\frac{3}{2}\)
\(=\frac{1}{2}\left(\frac{y}{x}+\frac{x}{y}+\frac{z}{x}+\frac{x}{z}+\frac{z}{y}+\frac{y}{z}\right)-\frac{3}{2}\ge\frac{3}{2}\)
\(=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\left(đpcm\right)\)
(BĐT cô-si trong ngoặc nha bạn nếu có sai cho mình xin lỗi!)
