Áp dụng BĐT AM-GM ta có:
\(\frac{2a}{2a+1}=1-\frac{1}{2a+1}\ge\frac{1}{2b+1}+\frac{1}{2c+1}\)\(\ge2\sqrt{\frac{1}{\left(2b+1\right)\left(2c+1\right)}}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\frac{2b}{2b+1}\ge2\sqrt{\frac{1}{\left(2a+1\right)\left(2c+1\right)}};\frac{2c}{2c+1}\ge2\sqrt{\frac{1}{\left(2a+1\right)\left(2b+1\right)}}\)
Nhân theo vế 3 BĐT trên ta có:
\(\frac{2a}{2a+1}\cdot\frac{2b}{2b+1}\cdot\frac{2c}{2c+1}\ge8\sqrt{\frac{1}{\left(2a+1\right)^2\left(2b+1\right)^2\left(2c+1\right)^2}}\)
\(\Leftrightarrow\frac{8abc}{\left(2a+1\right)\left(2b+1\right)\left(2c+1\right)}\ge\frac{8}{\left(2a+1\right)\left(2b+1\right)\left(2c+1\right)}\)
\(\Leftrightarrow8abc\ge8\Leftrightarrow abc\ge1\) (đúng)
Đẳng thức xảy ra khi \(a=b=c=1\)
