\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\) =>  (yz + xz + xy) / xyz = 0  => yz + zx + xy = 0 

Ta có : x² + 2yz = x² + yz + yz = x² + yz - zx - xy = x.(x - z) - y.(x - z) = (x - y).(x - z)

Tương tự, y² + 2xz = y² + xz + xz = y² + xz - xy - yz = y(y - x) + z(x - y) = (x - y)(z - y)

; z² + 2xy = (x - z).(y - z)

Vậy \(A=\frac{yz}{\left(x-y\right)\left(x-z\right)}+\frac{xz}{\left(x-y\right)\left(z-y\right)}+\frac{xy}{\left(x-z\right)\left(y-z\right)}\)

\(A=\frac{yz\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}-\frac{xz\left(x-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}+\frac{xy\left(x-y\right)}{\left(x-z\right)\left(y-z\right)\left(x-y\right)}\)

\(A=\frac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{yz\left(y-z\right)-xz\left(x-y+y-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)

\(A=\frac{\left(yz-xz\right)\left(y-z\right)+\left(xy-xz\right)\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\frac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)

cảm ơn

212121

đáp án

chia cả 2 vế của giả thiết cho xyz rồi đặt 1/x ; 1/y ; 1/z => a ; b ; c

đến đây thì tự làm tiếp đi 

Từ dữ kiện đề bài => x + y + z = xyz

Ta có : 

\(\frac{x}{\sqrt{yz\left(1+x^2\right)}}=\frac{x}{\sqrt{yz+xyz.x}}=\frac{x}{\sqrt{yz+x\left(x+y+z\right)}}=\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}\)

                                                                                                                   \(=\frac{\sqrt{x}}{\sqrt{x+z}}.\frac{\sqrt{x}}{\sqrt{x+y}}\le\frac{1}{2}.\left(\frac{x}{x+z}+\frac{x}{x+y}\right)\)

Tương tự với hai hạng tử còn lại , suy ra 

\(Q\le\frac{1}{2}\left(\frac{x}{x+z}+\frac{x}{x+y}\right)+\frac{1}{2}\left(\frac{y}{x+y}+\frac{y}{y+z}\right)+\frac{1}{2}\left(\frac{z}{z+x}+\frac{z}{z+y}\right)=\frac{3}{2}\)

Vậy Max = 3/2 <=> x = y = z 

Nguồn : Đinh Đức Hùng 

Từ \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)

\(\Rightarrow\)\(x+y+z=xyz\)

Ta có : \(\sqrt{yz\left(1+x^2\right)}=\sqrt{yz+x^2yz}=\sqrt{yz+x\left(x+y+z\right)}=\sqrt{\left(x+y\right)\left(x+z\right)}\)

Tương tự : \(\sqrt{xy\left(1+z^2\right)}=\sqrt{\left(z+y\right)\left(z+x\right)}\); \(\sqrt{zx\left(1+y^2\right)}=\sqrt{\left(y+z\right)\left(y+x\right)}\)

Nên \(Q=\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\frac{y}{\sqrt{\left(y+z\right)\left(y+x\right)}}+\frac{z}{\sqrt{\left(z+x\right)\left(z+y\right)}}\)

         \(Q=\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}+\sqrt{\frac{y}{x+y}.\frac{y}{y+z}}+\sqrt{\frac{z}{x+z}.\frac{z}{y+z}}\)

Áp dụng BĐT \(\sqrt{A.B}\le\frac{A+B}{2}\left(A,B>0\right)\)

Dấu "=" xảy ra khi A = B :

Ta được :

\(Q\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+x}+\frac{y}{y+z}+\frac{z}{z+x}+\frac{z}{z+y}\right)=\frac{3}{2}\)

Vậy GTLN của \(Q=\frac{3}{2}\)khi \(x=y=z=\sqrt{3}\)

Ta có: \(P=\frac{\sqrt{x}}{1+x+xy}+\frac{\sqrt{y}}{1+y+yz}+\frac{\sqrt{z}}{1+z+xz}\)

\(P=\frac{\sqrt{x}}{xy+x+1}+\frac{x\sqrt{y}}{x+xy+xyz}+\frac{xy\sqrt{z}}{xy+xyz+x^2yz}\)

\(P=\frac{\sqrt{x}}{xy+x+1}+\frac{x\sqrt{y}}{xy+x+1}+\frac{\sqrt{xy}.\sqrt{xyz}}{xy+x+1}\)

\(P=\frac{\sqrt{x}+x\sqrt{y}+\sqrt{xy}}{xy+x+1}\le\frac{\frac{x+1}{2}+\frac{x\left(y+1\right)}{2}+\frac{xy+1}{2}}{xy+x+1}\) (bđt cosi)

=> \(P\le\frac{x+1+xy+x+xy+1}{2\left(xy+x+1\right)}=\frac{2\left(xy+x+1\right)}{2\left(xy+x+1\right)}=1\)

Dấu "=" xảy ra<=> x =  y = z = 1

Vậy MaxP = 1 <=> x = y = z = 1