làm nhanh giùm mình nha ! đang cần gấp <:)

Sửa lại đề là x;y;z khác -1.

\(A=\frac{xy+2x+1}{xy+x+y+1}+\frac{yz+2y+1}{yz+y+z+1}+\frac{zx+2z+1}{zx+z+x+1}=\)

\(A=\frac{x\left(y+1\right)+x+1}{x\left(y+1\right)+y+1}+\frac{y\left(z+1\right)+y+1}{y\left(z+1\right)+z+1}+\frac{z\left(x+1\right)+z+1}{z\left(x+1\right)+x+1}=\)

\(A=\frac{x\left(y+1\right)+x+1}{\left(x+1\right)\left(y+1\right)}+\frac{y\left(z+1\right)+y+1}{\left(y+1\right)\left(z+1\right)}+\frac{z\left(x+1\right)+z+1}{\left(z+1\right)\left(x+1\right)}=\)vì x;y;z khác -1 nên:

\(A=\frac{x}{x+1}+\frac{1}{y+1}+\frac{y}{y+1}+\frac{1}{z+1}+\frac{z}{z+1}+\frac{1}{x+1}=\)

\(A=\frac{x}{x+1}+\frac{1}{x+1}+\frac{y}{y+1}+\frac{1}{y+1}+\frac{z}{z+1}+\frac{1}{z+1}=\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}=1+1+1=3\)

A = 3 với mọi x;y;z khác -1 nên A không phụ thuộc vào x;y;z. đpcm

Có \(VT=\dfrac{x^2}{x^3-xyz+2013x}+\dfrac{y^2}{y^3-xyz+2013y}+\dfrac{z^2}{z^3-xyz+2013z}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)

\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[x^2+y^2+z^2-\left(xy+yz+zx\right)\right]+2013\left(x+y+z\right)}\)

\(=\dfrac{x+y+z}{x^2+y^2+z^2-\left(xy+yz+zx\right)+3\left(xy+yz+zx\right)}\) 

(vì \(2013=3.671=3\left(xy+yz+zx\right)\))

\(=\dfrac{x+y+z}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}\)

\(=\dfrac{x+y+z}{\left(x+y+z\right)^2}\)

\(=\dfrac{1}{x+y+z}\)

ĐTXR \(\Leftrightarrow\dfrac{1}{x^2-yz+2013}=\dfrac{1}{y^2-zx+2013}=\dfrac{1}{z^2-xy+2013}\)

\(\Leftrightarrow x^2-yz=y^2-zx=z^2-xy\)

\(\Leftrightarrow x=y=z\) (với \(x,y,z>0\))

Vậy ta có đpcm.

Lời giải:

$2\text{VT}=2(x+y+z)-4(xy+yz+xz)+8xyz$

$=(2x-1)(2y-1)(2z-1)+1$

Do $x,y,z\in [0;1]$ nên $-1\leq 2x-1, 2y-1, 2z-1\leq 1$

$\Rightarrow (2x-1)(2y-1)(2z-1)\leq 1$

$\Rightarrow 2\text{VT}\leq 2$

$\Rightarrow \text{VT}\leq 1$
Ta có đpcm.

Dấu "=" xảy ra khi $(x,y,z)=(1,1,1), (0,0,1)$ và hoán vị.

\(=\dfrac{xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)}{xy\left(z+1\right)+y\left(z+1\right)-x\left(z+1\right)-\left(z+1\right)}\\ =\dfrac{\left(z-1\right)\left(xy-y-x+1\right)}{\left(z+1\right)\left(xy+y-x-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)\left(y-1\right)}{\left(z+1\right)\left(x+1\right)\left(y-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)}{\left(z+1\right)\left(x+1\right)}\\ =\dfrac{\left(5003-1\right)\left(5001-1\right)}{\left(5003+1\right)\left(5001+1\right)}=\dfrac{5002\cdot5000}{5004\cdot5002}=\dfrac{5000}{5004}=\dfrac{1250}{1251}\)