Ta có: A= \(\dfrac{xy+2y+1}{xy+x+y+1}+\dfrac{yz+2z+1}{yz+y+z+1}\) +\(\dfrac{zx+2x+1}{zx+z+x+1}\)

=\(\dfrac{xy+2y+1}{\left(x+1\right)\left(y+1\right)}+\dfrac{yz+2z+1}{\left(y+1\right)\left(z+1\right)}\) +\(\dfrac{zx+2x+1}{\left(x+1\right)\left(z+1\right)}\)

=\(\dfrac{\left(xy+2y+1\right)\left(z+1\right)}{\left(z+1\right)\left(y+1\right)\left(x+1\right)}\)+\(\dfrac{\left(yz+2z+1\right)}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)+\(\dfrac{\left(y+1\right)\left(zx+2x+1\right)}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

Đặt B =(z+1)(xy+2y+1)+(yz+2z+1)(x+1)+(y+1)(zx+2x+1)

=>B= xyz+2yz+z+xy+2y+1+xyz+2zx+x+yz+2z+1+xyz+2xy+y+xz+2x+1 = 3xyz+3yz+3z+3xy+3y+3+3xz+3x = 3(xyz+yz +x+1+xy+y+xz+z) =3[yz(x+1)+(x+1)+y(x+1)+z(x+1)] =3(x+1)(yz+y+z+1)=3(x+1)(y+1)(1+z)

=> A=\(\dfrac{B}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)=\(\dfrac{3\left(x+1\right)\left(y+1\right)\left(z+1\right)}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)=3

Vậy A=3 với mọi x,y,z

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

Ta có: \(\frac{xy+2x+1}{xy+x+y+1}=\frac{\left(xy+x\right)+\left(x+1\right)}{\left(xy+x\right)+\left(y+1\right)}=\frac{x\left(y+1\right)+\left(x+1\right)}{\left(y+1\right)\left(x+1\right)}=\frac{x}{x+1}+\frac{1}{y+1}\)

Tương tự ta có:

\(\frac{yz+2y+1}{yz+y+z+1}=\frac{y}{y+1}+\frac{1}{z+1}\)

\(\frac{zx+2z+1}{zx+z+x+1}=\frac{z}{z+1}+\frac{1}{x+1}\)

Từ đây ta có biểu thức ban đầu sẽ bằng

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

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

CHÚ Ý: ab+a+b+1=a(b+1)+(b+1)=(a+1)(b+1)

Xét: \(\frac{xy+2x+1}{xy+x+y+1}=\frac{x\left(y+1\right)+x+1}{\left(x+1\right)\left(y+1\right)}=\frac{x}{x+1}+\frac{1}{y+1}\)

Tương tự với 2 biểu thức còn lại ta được:

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

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

?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????

Sửa lại đề: cho x, y, z dương thỏa mãn \(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}=1\)

Chứng minh \(A=\dfrac{x}{\sqrt{yz\left(1+x^2\right)}}+\dfrac{y}{\sqrt{xz\left(1+y^2\right)}}+\dfrac{z}{\sqrt{xy\left(1+z^2\right)}}\le\dfrac{3}{2}\)

Giải:

Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac{1}{z}\Rightarrow ab+bc+ac=1\)

\(\Rightarrow A=\dfrac{\dfrac{1}{a}}{\sqrt{\dfrac{1}{bc}\left(1+\dfrac{1}{a^2}\right)}}+\dfrac{\dfrac{1}{b}}{\sqrt{\dfrac{1}{ac}\left(1+\dfrac{1}{b^2}\right)}}+\dfrac{\dfrac{1}{a}}{\sqrt{\dfrac{1}{ab}\left(1+\dfrac{1}{c^2}\right)}}\)

\(\Rightarrow A=\sqrt{\dfrac{bc}{a^2+1}}+\sqrt{\dfrac{ac}{b^2+1}}+\sqrt{\dfrac{ab}{c^2+1}}\)

\(\Rightarrow A=\sqrt{\dfrac{bc}{a^2+ab+bc+ac}}+\sqrt{\dfrac{ac}{b^2+ab+bc+ac}}+\sqrt{\dfrac{ab}{c^2+ab+bc+ac}}\)

\(\Rightarrow A=\sqrt{\dfrac{bc}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{ac}{\left(a+b\right)\left(b+c\right)}}+\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\)

\(\Rightarrow A\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}+\dfrac{a}{a+b}+\dfrac{c}{b+c}+\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)

\(\Rightarrow A\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{a+c}{a+c}\right)=\dfrac{3}{2}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{\sqrt{3}}{3}\) hay \(x=y=z=\sqrt{3}\)

Đề bài này có rất nhiều vấn đề, đầu tiên không có điều kiện x, y, z gì cả? Dương? Â? Bằng 0? Khác 0?

Sau nữa là chiều của BĐT cũng có vấn đề nốt, mình thử với \(x=y=2;z=\dfrac{4}{3}\) thì vế trái ra \(\dfrac{2+\sqrt{30}}{5}\) mà theo casio cho biết thì số này nhỏ hơn \(\dfrac{3}{2}\) , vậy BĐT cũng sai luôn

Do \(xyz\ne0\) ta có:

\(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}=0\Leftrightarrow xyz\left(\dfrac{1}{xy}+\dfrac{1}{xz}+\dfrac{1}{yz}\right)=0\Leftrightarrow x+y+z=0\)

Lại có: \(x^3+y^3+z^3=x^3+y^3+3x^2y+3y^2x-3xy\left(x+y\right)+z^3\)

\(=\left(x+y\right)^3+z^3-3xy\left(-z\right)=\left(x+y+z\right)\left(\left(x+y\right)^2-\left(x+y\right)z+z^2\right)+3xyz=3xyz\)

Vậy nếu \(x+y+z=0\) thì \(x^3+y^3+z^3=3xyz\)

\(P=\dfrac{x^2}{yz}+\dfrac{y^2}{xz}+\dfrac{z^2}{xy}=\dfrac{x^3}{xyz}+\dfrac{y^3}{xyz}+\dfrac{z^3}{xyz}=\dfrac{x^3+y^3+z^3}{xyz}=\dfrac{3xyz}{xyz}=3\)