\(BDT\Leftrightarrow\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\ge1\)
Do abc=1 nên tồn tại \(\left(a,b,c\right)~\left(\frac{x}{y};\frac{y}{z};\frac{z}{x}\right)\)
thay vào,\(BDT\Leftrightarrow\frac{x}{x+2y}+\frac{y}{y+2z}+\frac{z}{z+2x}\ge1\)
Áp dụng BĐT cauchy-schwarz:
\(\frac{x^2}{x^2+2xy}+...\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\left(đPcM\right)\)