đặt [tex](a;b;c)=(\frac{1}{x};\frac{1}{y};\frac{1}{z})[/tex]
Từ gt => xy+yz+zx=6
Ta cần cm: [tex]\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}\geq 2[/tex]
Ta có: [tex]VT=\sum \frac{x^4}{yx+2zx}\geq \frac{(x^2+y^2+z^2)^2}{3(xy+yz+zx)}[/tex] (bđt CBS)
[tex]VT\geq \frac{(xy+yz+zx)^2}{3(xy+yz+zx)}=2[/tex] (đpcm)
đẳng thức xảy ra <=> [tex]x=y=z=\sqrt{2}\Leftrightarrow a=b=c=\frac{1}{\sqrt{2}}[/tex]