Chứng minh a/(a+1)+b/(b+1)+c/(c+1) < = 3/4

Xét: \(\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\)

\(\Leftrightarrow\frac{a}{2a+b+c}+\frac{b}{a+2b+c}+\frac{c}{a+b+2c}\)

Áp dụng bất đẳng thức \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\) với a , b > 0

\(\Rightarrow\left\{\begin{matrix}\frac{a}{2a+b+c}=\frac{a}{a+b+a+c}\le\frac{a}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\\\frac{b}{a+2b+c}=\frac{b}{a+b+b+c}\le\frac{b}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)\\\frac{c}{a+b+2c}=\frac{c}{a+c+b+c}\le\frac{c}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\end{matrix}\right.\)

\(\Rightarrow VT\le\frac{a}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)+\frac{b}{4}\left(\frac{1}{a+b}+\frac{1}{b+c}\right)+\frac{c}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\)

\(\Rightarrow VT\le\frac{a}{4\left(a+b\right)}+\frac{a}{4\left(a+c\right)}+\frac{b}{4\left(a+b\right)}+\frac{b}{4\left(b+c\right)}+\frac{c}{4\left(a+c\right)}+\frac{c}{4\left(b+c\right)}\)

\(\Rightarrow VT\le\left[\frac{a}{4\left(a+b\right)}+\frac{b}{4\left(a+b\right)}\right]+\left[\frac{a}{4\left(a+c\right)}+\frac{c}{4\left(a+c\right)}\right]+\left[\frac{b}{4\left(b+c\right)}+\frac{c}{4\left(b+c\right)}\right]\)

\(\Rightarrow VT\le\frac{a+b}{4\left(a+b\right)}+\frac{a+c}{4\left(a+c\right)}+\frac{b+c}{4\left(b+c\right)}\)

\(\Rightarrow VT\le\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\)

\(\Rightarrow VT\le\frac{3}{4}\)

\(\Leftrightarrow\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\le\frac{3}{4}\) ( đpcm )