$$M=\frac{x^2}{x^2+yz+x+1}+\frac{y+z}{z+y+x+1}+\frac{1}{xyz+3}$$
Lời giải: Ta sẽ CM :$M\leq 1$
Ta có $2(1-xy-xz+yz)=x^2+y^2+z^2-2xy-2xz+2yz=(x-y-z)^2\geq 0$
$= > x^2+yz+x+1=x(x+y+z+1)+(1-xy-xz+yz)\geq x(x+y+z+1)$
$= > \dfrac{x^2}{x^2+yz+x+1}\leq \dfrac{x^2}{x(x+y+z+1)}=\dfrac{x}{x+y+z+1}$
$= > \dfrac{x^2}{x^2+yz+x+1}\leq \dfrac{x^2}{x(x+y+z+1)}=\dfrac{x}{x+y+z+1}$
$= > M\leq \dfrac{x}{x+y+z+1}+\dfrac{y+z}{x+y+z+1}+\dfrac{1}{xyz+3}=1-\dfrac{1}{x+y+z+1}+\dfrac{1}{xyz+3}\leq 1$
$< = > \dfrac{1}{x+y+z+1}\geq \dfrac{1}{xyz+3}$
$< = > xyz+2\geq x+y+z$
$< = > \dfrac{1}{x+y+z+1}\geq \dfrac{1}{xyz+3}$
$< = > xyz+2\geq x+y+z$
Theo bđt Bunhiacopxki có $\left [ x+y+z-xyz \right ]^2=\left [ x(1-yz)+(y+z) \right ]^2\leq (x^2+(y+z)^2)((1-yz)^2+1^2)=(2+2yz)(y^2z^2-2yz+2)\leq 4$
$< = > (yz)^2(yz-1)\leq 0< = > yz\leq 1$
$< = > (yz)^2(yz-1)\leq 0< = > yz\leq 1$
Nhưng bđt đúng vì $2=x^2+y^2+z^2\geq y^2+z^2\geq 2yz= > yz\leq 1$
Do đó ta có ĐPCM .Dấu = xảy ra khi $x=0,y=z=1$
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