đặt $b+c=x;a+c=y;a+b=z$$VT=\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}=\frac{1}{2}[\frac{y}{x}+\frac{x}{y}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}-3]$Ta có:$\frac{x}{y}+\frac{y}{z}\geq2 $ ( cauchy)$\Rightarrow VT\geq \frac{1}{2}(6-3)=\frac{3}{2}$"=" tại $a=b=c$
đặt $b+c=x;a+c=y;a+b=z$$VT=\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}=\frac{1}{2}[\frac{y}{x}+\frac{x}{y}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}-3]$$\frac{x}{y}+\frac{y}{z}\geq2 $ ( cauchy)$\Rightarrow VT\geq \frac{1}{2}(6-3)=\frac{3}{2}$"=" tại $a=b=c$
đặt $b+c=x;a+c=y;a+b=z$$VT=\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}=\frac{1}{2}[\frac{y}{x}+\frac{x}{y}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}-3]$
Ta có:$\frac{x}{y}+\frac{y}{z}\geq2 $ ( cauchy)$\Rightarrow VT\geq \frac{1}{2}(6-3)=\frac{3}{2}$"=" tại $a=b=c$