\(Cho\)\(x,y,z\)\(khác\)\(0\)\(thỏa\)\(mãn\)

\(\hept{\begin{cases}x+y+z=\frac{1}{2}\\\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}>0\end{cases}}+\frac{1}{xyz}=4\)          \(Tính\)\(\left(x^{2013}+y^{2013}\right)\left(y^{2015}+z^{2015}\right)\left(z^{2015}+x^{2015}\right)\)

Bài này hơi khó nha giúp mình

cho \(x;y;z>0\)

\(xy+yz+xz=xyz\)

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

tính giá trị của biểu thức

\(A=\sqrt{\frac{\left(2x+yz\right)\left(2y+xz\right)}{\left(y+z\right)\left(x+z\right)}}+\sqrt{\frac{\left(2y+xz\right)\left(2z+xy\right)}{\left(x+z\right)\left(x+y\right)}}+\sqrt{\frac{\left(2z+xy\right)\left(2x+yz\right)}{\left(x+y\right)\left(y+z\right)}}\)

\(\hept{\begin{cases}x+\frac{1}{x}=y^2+1\left(1\right)\\y+\frac{1}{y}=z^2+1\left(2\right)\\z+\frac{1}{z}=x^2+1\left(3\right)\end{cases}}\left(ĐKXĐ:x;y;z\ne0\right)\)

Từ \(\left(1\right)\Rightarrow\frac{x^2+1}{x}=y^2+1\)

              \(\Rightarrow x=\frac{x^2+1}{y^2+1}>0\)

Tương tự \(y=\frac{y^2+1}{z^2+1}>0\)

                \(z=\frac{z^2+1}{x^2+1}>0\)

Áp dụng bđt cô-si có \(x+\frac{1}{x}\ge2\) 

\(\Rightarrow y^2+1\ge2\)

\(\Rightarrow y\ge1\)

Tương tự \(x;z\ge1\)

ta có \(xyz=\frac{\left(x^2+1\right)\left(y^2+1\right)\left(z^2+1\right)}{\left(x^2+1\right)\left(y^2+1\right)\left(z^2+1\right)}=1\)

Mà \(x;y;z\ge1\Rightarrow xyz\ge1\)

Do đó dấu "=" khi x = y = z = 1 

Yey =)))

chị QA

ta có đề bài <=> 

\(\frac{x^2}{y}-2x+y+\frac{y^2}{z}-2y+z+\frac{z^2}{x}-2z+x+\left(x+y+z\right)-\left(x-y\right)^2-\left(y-z\right)^2-\left(z-x\right)^2\)

=\(\frac{\left(x-y\right)^2}{y}-\left(x-y\right)^2+...+\left(x+y+z\right)\)

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

mà \(\sqrt{x}+\sqrt{y}+\sqrt{z}=1\Rightarrow x,y,z\in\left[0;1\right]\)

=> \(\frac{1}{y}-y>0\)

=> \(A\ge x+y+z\ge\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{3}=\frac{1}{3}\)