\(\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge2\)
\(\Leftrightarrow\dfrac{a}{a+1}\ge2-\dfrac{b}{b+1}-\dfrac{c}{c+1}\)
\(\Leftrightarrow\dfrac{a}{a+1}\ge2-\dfrac{b}{b+1}-\dfrac{c}{c+1}\)
\(\Leftrightarrow\dfrac{a}{a+1}\ge\left(1-\dfrac{b}{b+1}\right)+\left(1-\dfrac{c}{c+1}\right)\)
\(\Leftrightarrow\dfrac{a}{a+1}\ge\dfrac{1}{b+1}+\dfrac{1}{c+1}\)
- Áp dụng bất đẳng thức Caushy cho các số dương:
\(\dfrac{1}{b+1}+\dfrac{1}{c+1}\ge\dfrac{2}{\sqrt{\left(b+1\right)\left(c+1\right)}}\)
\(\Rightarrow\dfrac{a}{a+1}\ge\dfrac{2}{\sqrt{\left(b+1\right)\left(c+1\right)}}>0\left(1\right)\)
- Tương tự:
\(\dfrac{b}{b+1}\ge\dfrac{2}{\sqrt{\left(c+1\right)\left(a+1\right)}}>0\left(2\right)\)
\(\dfrac{c}{c+1}\ge\dfrac{2}{\sqrt{\left(a+1\right)\left(b+1\right)}}>0\left(3\right)\)
\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow\dfrac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{8}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\)
\(\Rightarrow abc\ge8\)
- Dấu "=" xảy ra \(\Leftrightarrow a=b=c=2\)
