Lời giải:
BĐT cần chứng mình tương đương với:

$(xy+yz+xz)^2\geq 3(x+y+z)$

$\Leftrightarrow (xy+yz+xz)^2\geq 3xyz(x+y+z)$

$\Leftrightarrow (xy)^2+(yz)^2+(zx)^2+2xyz(x+y+z)\geq 3xyz(x+y+z)$

$\Leftrightarrow (xy)^2+(yz)^2+(xz)^2\geq xyz(x+y+z)$

$\Leftrightarrow (xy)^2+(yz)^2+(xz)^2-xyz(x+y+z)\geq 0$

$\Leftrightarrow 2(xy)^2+2(yz)^2+2(xz)^2-2xyz(x+y+z)\geq 0$

$\Leftrightarrow (xy-yz)^2+(yz-xz)^2+(xz-xy)^2\geq 0$

(luôn đúng với mọi $x,y,z\geq 0$)

Dấu "=" xảy ra khi $x=y=z=1$

\(M=\frac{x}{xy+x+2015}+\frac{y}{yz+y+1}+\frac{2015z}{xz+2015z+2015}\)

\(\Leftrightarrow M=\frac{x}{xy+x+xyz}+\frac{y}{yz+y+1}+\frac{xyz.z}{xz+xyz.z+xyz}\left(xyz=2015\right)\)

\(\Leftrightarrow M=\frac{1}{y+1+yz}+\frac{y}{yz+y+1}+\frac{yz}{1+yz+y}\)

\(\Leftrightarrow M=\frac{yz+y+1}{yz+y+1}=1\)

\(M=\frac{x}{xy+x+2015}+\frac{y}{yz+y+1}+\frac{2015z}{xz+2015z+2015}\)

Thay xyz = 2015, Ta có: 

\(M=\frac{x}{xy+x+xyz}+\frac{y}{yz+y+1}+\frac{xyz^2}{xz+xyz^2+xyz}\)

\(M=\frac{1}{y+1+yz}+\frac{y}{yz+y+1}+\frac{yz}{1+yz+y}\)

\(M=\frac{y+1+yz}{y+1+yz}=1\)

Bạn viết đề rõ ràng hơn nhé, mình không đọc được :(

mik đăng cái khác rồi đó

khó đọc

Câu 1: 

\(a^2+b^2-a^2b^2+ab-a-b\)

\(=a^2\left(1-b^2\right)+b\left(b-1\right)+a\left(b-1\right)\)

\(=-a^2\left(b-1\right)\left(b+1\right)+\left(b-1\right)\left(a+b\right)\)

\(=\left(b-1\right)\left(-a^2b-a^2+a+b\right)\)

\(=\left(b-1\right)\cdot\left[-b\left(a^2-1\right)-a\left(a-1\right)\right]\)

\(=\left(b-1\right)\left(a-1\right)\left[-b\left(a+1\right)-a\right]\)