Question

cho   \(x+y+z=1\)

tim max \(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)

Answer 1

\(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)

\(=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)

\(\le3-\frac{9}{3+x+y+z}=3-\frac{9}{3+1}=\frac{3}{4}\)

Answer 2

= 1+x+1+y+1+z

= 3+x+y+z

=3+1=4

Answer 3

\(\frac{3}{4}\)nha

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