Lời giải:
Vì $0\leq x\leq y\leq z\leq 1\Rightarrow 0\leq xy\leq xz\leq yz$
$\Rightarrow \frac{x}{yz+1}+\frac{y}{xz+1}+\frac{z}{xy+1}\leq \frac{x+y+z}{xy+1}(1)$
Xét $\frac{x+y+z}{xy+1}-2=\frac{x+y+z-2xy-2}{xy+1}=\frac{(x-1)(1-y)+(z-xy-1)}{xy+1}\leq 0$ do $0\leq x\leq y\leq z\leq 1$)
$\Rightarrow \frac{x+y+z}{xy+1}\leq 2(2)$
Từ $(1);(2)\Rightarrow \frac{x}{yz+1}+\frac{y}{xz+1}+\frac{z}{xy+1}\leq 2$ (đpcm)
Lời giải:
Vì 0≤x≤y≤z≤1⇒0≤xy≤xz≤yz0≤x≤y≤z≤1⇒0≤xy≤xz≤yz
⇒xyz+1+yxz+1+zxy+1≤x+y+zxy+1(1)⇒xyz+1+yxz+1+zxy+1≤x+y+zxy+1(1)
Xét x+y+zxy+1−2=x+y+z−2xy−2xy+1=(x−1)(1−y)+(z−xy−1)xy+1≤0x+y+zxy+1−2=x+y+z−2xy−2xy+1=(x−1)(1−y)+(z−xy−1)xy+1≤0 do 0≤x≤y≤z≤10≤x≤y≤z≤1)
⇒x+y+zxy+1≤2(2)⇒x+y+zxy+1≤2(2)
Từ (1);(2)⇒xyz+1+yxz+1+zxy+1≤2(1);(2)⇒xyz+1+yxz+1+zxy+1≤2 (đpcm)
