Bđt $\Leftrightarrow \frac{\sum xy}{x+z} \geq \frac{y^{2}+yz+z^{2}}{y+z} $ $\Leftrightarrow \frac{\sum xy}{x+z}-y \geq \frac{y^{2}+yz+z^{2}}{y+z}-y$$\Leftrightarrow zx(y+z)-z^{2}(x+z)\geq 0$$\Leftrightarrow z(xy-z^{2}) \geq 0 \rightarrow (đúng ) (x\geq y\geq z\geq 0)$Dấu = xảy ra $\Leftrightarrow x=y=z$ hoặc $x=y,z=0$
Bđt $\Leftrightarrow \frac{\sum xy}{x+z} \geq \frac{y^{2}+yz+z^{2}}{y+z} $ $\Leftrightarrow \frac{\sum xy}{x+z}-y \geq \frac{y^{2}+yz+z^{2}}{y+z}$$\Leftrightarrow zx(y+z)-z^{2}(x+z)\geq 0$$\Leftrightarrow z(xy-z^{2}) \geq 0 \rightarrow (đúng ) (x\geq y\geq z\geq 0)$Dấu = xảy ra $\Leftrightarrow x=y=z$ hoặc $x=y,z=0$
Bđt $\Leftrightarrow \frac{\sum xy}{x+z} \geq \frac{y^{2}+yz+z^{2}}{y+z} $ $\Leftrightarrow \frac{\sum xy}{x+z}-y \geq \frac{y^{2}+yz+z^{2}}{y+z}
-y$$\Leftrightarrow zx(y+z)-z^{2}(x+z)\geq 0$$\Leftrightarrow z(xy-z^{2}) \geq 0 \rightarrow (đúng ) (x\geq y\geq z\geq 0)$Dấu = xảy ra $\Leftrightarrow x=y=z$ hoặc $x=y,z=0$