Question

cho x,y,z là số nguyên dương và x+y+z=1 tìm max của

\(P=\dfrac{xy}{z+1}+\dfrac{yz}{x+1}+\dfrac{xz}{y+1}\)

Answer 1

Ta có:\(\left(x+y\right)^2\ge4xy\)(tự cm)

\(\Rightarrow xy\le\dfrac{\left(x+y\right)^2}{4}\)

CMTT\(\Rightarrow yz\le\dfrac{\left(y+z\right)^2}{4};xz\le\dfrac{\left(x+z\right)^2}{4}\)

\(\Rightarrow P\le\dfrac{\left(x+y\right)^2}{4z+4}+\dfrac{\left(y+z\right)^2}{4x+4}+\dfrac{\left(x+z\right)^2}{4y+4}\)

\(\Rightarrow P\le\dfrac{\left(x+y+y+z+z+x\right)^2}{4x+4y+4z+4+4+4}=\dfrac{2^2}{16}=\dfrac{1}{4}\)

\(\Rightarrow MAXP=\dfrac{1}{4}\Leftrightarrow x=y=z=\dfrac{1}{3}\)