\(S=\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=1-\frac{1}{a+1}+1-\frac{1}{b+1}+1-\frac{1}{c+1}\)
\(=3-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\)
\(\frac{1}{a+1}+\frac{9}{16}\left(a+1\right)\ge2\sqrt{\frac{1}{\left(a+1\right)}\cdot\frac{9}{16}\left(a+1\right)}=2\cdot\frac{3}{4}=\frac{3}{2}\)
Tương tự \(\frac{1}{b+1}+\frac{9}{16}\left(b+1\right)\ge\frac{3}{2}\) ; \(\frac{1}{c+1}+\frac{9}{16}\left(c+1\right)\ge\frac{3}{2}\)
=> \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{9}{16}\left(a+b+c+3\right)\ge\frac{9}{2}\)
=> \(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}+\frac{9}{4}\ge\frac{9}{2}\Leftrightarrow\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{9}{4}\)
\(\Rightarrow S=3-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\le3-\frac{9}{4}=\frac{3}{4}\)
Vậy MAX S= 3/4 . Dấu '' = '' xảy ra khi \(\frac{1}{a+1}=\frac{9}{16}\left(a+1\right);\frac{1}{b+1}=\frac{9}{16}\left(b+1\right);\frac{1}{c+1}=\frac{9}{16}\left(c+1\right)\)
a + b +c = 1
=> a = b =c =1/3
\(S=3-\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)\le3-\frac{9}{a+b+c+1+1+1}=3-\frac{9}{4}=\frac{3}{4}\)
