= có x +y+z=a=>x²+y²+z²+2(xy+yz+xz)=a²

Thay vào a²=b+3992=>xy+zy+xz=1996

thay vào P ta có

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

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

+\(\sqrt{\dfrac{\left(zx+xy+zy+x^2\right)\left(xz+xy+zy+y^2\right)}{xz+xy+zy+z^2}}\)

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

+y\(\sqrt{\dfrac{\left(x+z\right)\left(z+y\right)\left(x+y\right)\left(z+x\right)}{\left(y+z\right)\left(x+y\right)}}\)

+z\(\sqrt{\dfrac{\left(y+z\right)\left(x+y\right)\left(z+x\right)\left(x+y\right)}{\left(z+x\right)\left(z+y\right)}}\)

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

=2(xy+zx+zy)=3992

*có gì ko hiểu thì hỏi

x+y+z=0

nên x+y=-z; y+z=-x; x+z=-y

\(\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)

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

Chứng minh gì bạn?

Đề bài sai

Ví dụ: với \(a=1;b=2;c=3,d=4\) thì \(x=\dfrac{1}{2}\) ; \(y=\dfrac{3}{4}\) ; \(z=\dfrac{2}{3}\)

Khi đó  \(x< y\) nhưng \(z< y\)

\(\text{Vì }\dfrac{a}{b}< \dfrac{c}{d}\text{ nên }ad< bc\left(1\right)\)

\(\text{Xét tích}:a\left(b+d\right)=ab+ad\left(2\right)\)

                \(b\left(a+c\right)=ba+bc\left(3\right)\)

\(\text{Từ(1);(2);(3)}\Rightarrow a\left(b+d\right)< b\left(a+c\right)\text{ do đó }\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(4\right)\)

\(\text{Tương tự ta có:}\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(5\right)\)

\(\text{Từ (4);(5) ta được }\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)

\(\Rightarrow x< y< z\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

(y + z - x)/x = (z + x - y)/y = (x + y - z)/z = 1

--> y + z - x = x; z + x - y = y; x + y - z = z

--> y + z = 2x; z + x = 2y; x + y = 2z

Ta có: 

B = (x + y)/y.(y + z)/z.(z + x)/x

= 2z/y.2x/z.2y/x = 8