[TEX] cho x,y,z >0.Tim GTNN:
A=\frac{3x}{y+z}+\frac{4y}{z+x}+\frac{5z}{x+y} [/TEX]
[TEX]A+12=\frac{3x}{y+z}+3 +\frac{4y}{z+x}+4+\frac{5z}{x+y}+5[/TEX]
[TEX]A+12=\frac{3(x+y+z)}{y+z} + \frac{4(x+y+z)}{z+x} + \frac{5(x+y+z)}{x+y}[/TEX]
[TEX]A+12=\frac{1}{2}(y+z+z+x+x+y)(\frac{3}{y+z} + \frac{4}{z+x} + \frac{5}{x+y}) \geq \frac{1}{2}(\sqrt{3}+2+\sqrt{5})^2[/TEX](theo BĐT cauchy schwarz)
[TEX]A \geq \frac{1}{2}(\sqrt{3} + 2 \sqrt{5})^2 - 12[/TEX]
Dấu bằng xảy ra [TEX]\Leftrightarrow \frac{sqrt{3}}{y+z}=\frac{2}{z+x}=\frac{\sqrt{5}}{x+y}[/TEX]