Đáp án: $M\le \dfrac14$
Giải thích các bước giải:
Ta có:
$M=\dfrac{xy}{z+1}+\dfrac{yz}{x+1}+\dfrac{zx}{y+1}$
$\to M=\dfrac{xy}{z+x+y+z}+\dfrac{yz}{x+x+y+z}+\dfrac{zx}{y+x+y+z}$ vì $x+y+z=1$
$\to M=xy\cdot \dfrac{1}{\left(z+x\right)+\left(z+y\right)}+yz\cdot \dfrac{1}{\left(x+y\right)+\left(x+z\right)}+zx\cdot \dfrac{1}{\left(y+x\right)+\left(y+z\right)}$
$\to M\le xy\cdot \dfrac{1}{4}\left(\dfrac{1}{z+x}+\dfrac{1}{z+y}\right)+yz\cdot\dfrac14 \left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)+zx\cdot \dfrac14\left(\dfrac{1}{y+x}+\dfrac{1}{y+z}\right)$
$\to M\le \dfrac14\left(\dfrac{xy}{z+x}+\dfrac{xy}{z+y}+\dfrac{yz}{x+y}+\dfrac{yz}{x+z}+\dfrac{zx}{y+x}+\dfrac{zx}{y+z}\right)$
$\to M\le \dfrac14\left(\dfrac{xy+yz}{z+x}+\dfrac{xy+zx}{z+y}+\dfrac{yz+zx}{x+y}\right)$
$\to M\le \dfrac14\left(\dfrac{y\left(x+z\right)}{z+x}+\dfrac{x\left(y+z\right)}{z+y}+\dfrac{z\left(x+y\right)}{x+y}\right)$
$\to M\le \dfrac14\left(y+x+z\right)$
$\to M\le \dfrac14$ vì $x+y+z=1$
Dấu = xảy ra khi $x=y=z=\dfrac13$
