BĐT cần chứng minh tương đương:
\(\Leftrightarrow\frac{2}{2+a}+\frac{2}{2+b}+\frac{2}{2+c}\le2\)
\(\Leftrightarrow\frac{2}{2+a}-1+\frac{2}{2+b}-1+\frac{2}{2+c}-1\le2-3\)
\(\Leftrightarrow\frac{a}{a+2}+\frac{b}{b+2}+\frac{c}{c+2}\ge1\) (1)
Ta cần chứng minh (1)
Do \(abc=1\) nên tồn tại x;y;z sao cho: \(\left(a;b;c\right)=\left(\frac{x}{y};\frac{y}{z};\frac{z}{x}\right)\)
\(VT=\frac{x}{x+2y}+\frac{y}{y+2z}+\frac{z}{z+2x}=\frac{x^2}{x^2+2xy}+\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2xz}\)
\(VT\ge\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2xy+2yz+2xz}=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
