$1/$BĐT: $\Leftrightarrow y\frac{x+z}{xz}+\frac{x+z}{y}\leq \frac{\left ( x +z \right )^{2}}{xz}$$\Leftrightarrow y^{2}+xz\leq y\left ( x +z \right )$ ( vì $x, y, z>0$)$\Leftrightarrow y^{2}-\left ( x +z \right )y+xz\leq 0$$\Leftrightarrow \left ( y-x\right ) \left ( y-z\right )\leq 0$, vì $\begin{cases}y-x\geq 0 \\ y-z\leq 0\end{cases}$Nên BĐT cuối luôn đúng $\Rightarrow $ (ĐPCM)$2/$Ta có:$\begin{cases}\left ( 1-\frac{x}{y} \right )\left ( 1-\frac{y}{z} \right )\geq 0\\ \left ( 1-\frac{y}{x} \right )\left ( 1-\frac{z}{y} \right )\geq 0 \end{cases}\Rightarrow \frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}\leq 2+\frac{x}{z}+\frac{z}{x}$$\Rightarrow $ VT $= 3+ \frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}$ $\leq 5+2\left ( \frac{x}{z}+\frac{z}{x}\right )\leq 10+\left ( \frac{x}{z}-2\right )\left ( 2-\frac{z}{x}\right )\leq 10$ (vì $\frac{x}{z}, \frac{z}{x} \leq 2)$$\Rightarrow $ ĐPCM
$1/$BĐT: $\Leftrightarrow y\frac{x+z}{xz}+\frac{x+z}{y}\leq \frac{\left ( x +z \right )^{2}}{xz}$$\Leftrightarrow y^{2}+xz\leq y\left ( x +z \right )$ ( vì $x.y.z>0$)$\Leftrightarrow y^{2}-\left ( x +z \right )y+xz\leq 0$$\Leftrightarrow \left ( y-x\right ) \left ( y-z\right )\leq 0$, vì $\begin{cases}y-x\geq 0 \\ y-z\geq 0\end{cases}$Nên BĐT cuối luôn đúng $\Rightarrow $ (ĐPCM)$2/$Ta có:$\begin{cases}\left ( 1-\frac{x}{y} \right )\left ( 1-\frac{y}{z} \right )\geq 0\\ \left ( 1-\frac{y}{x} \right )\left ( 1-\frac{z}{y} \right )\geq 0 \end{cases}\Rightarrow \frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}\leq 2+\frac{x}{z}+\frac{z}{x}$$\Rightarrow $ VT $= 3+ \frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}$$\leq 5+2\left ( \frac{x}{z}+\frac{z}{x}\right )\leq 10+\left ( \frac{x}{z}-2\right )\left ( 2-\frac{z}{x}\right )\leq 10$
$1/$BĐT: $\Leftrightarrow y\frac{x+z}{xz}+\frac{x+z}{y}\leq \frac{\left ( x +z \right )^{2}}{xz}$$\Leftrightarrow y^{2}+xz\leq y\left ( x +z \right )$ ( vì $x
, y
, z>0$)$\Leftrightarrow y^{2}-\left ( x +z \right )y+xz\leq 0$$\Leftrightarrow \left ( y-x\right ) \left ( y-z\right )\leq 0$, vì $\begin{cases}y-x\geq 0 \\ y-z\
leq 0\end{cases}$Nên BĐT cuối luôn đúng $\Rightarrow $ (ĐPCM)$2/$Ta có:$\begin{cases}\left ( 1-\frac{x}{y} \right )\left ( 1-\frac{y}{z} \right )\geq 0\\ \left ( 1-\frac{y}{x} \right )\left ( 1-\frac{z}{y} \right )\geq 0 \end{cases}\Rightarrow \frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}\leq 2+\frac{x}{z}+\frac{z}{x}$$\Rightarrow $ VT $= 3+ \frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}+\frac{x}{z}$
$\leq 5+2\left ( \frac{x}{z}+\frac{z}{x}\right )\leq 10+\left ( \frac{x}{z}-2\right )\left ( 2-\frac{z}{x}\right )\leq 10$
(vì $\frac{x}{z}, \frac{z}{x} \leq 2)$$\Rightarrow $ ĐPCM