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\(\frac{2}{x^2+y^2}+\frac{2}{y^2+z^2}+\frac{2}{z^2+x^2}=3+\frac{z^2}{x^2+y^2}+\frac{x^2}{y^2+z^2}+\frac{y^2}{z^2+x^2}\le3+\frac{z^2}{2xy}+\frac{x^2}{2yz}+\frac{y^2}{2zx}\)
\(=3+\frac{x^3+y^3+z^3}{2xyz}\)
\(\Rightarrow\)\(A\le3\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=\sqrt{\frac{2}{3}}\)
1, A= y^3(1-y)^2 = 4/9 . y^3 . 9/4 (1-y)^2
= 4/9 .y.y.y . (3/2-3/2.y)^2
=4/9 .y.y.y (3/2-3/2.y)(3/2-3/2.y)
<= 4/9 (y+y+y+3/2-3/2.y+3/2-3/2.y)^5
=4/9 . 243/3125
=108/3125
Đến đó tự giải
Cộng vế theo vế
=> \(x^2+x+y^2+y+z^2+z=x^2+y^2+z^2\)
=> \(x+y+z=0\)=> A = 0
\(x=\left(y^2-x^2\right)=\left(y-x\right)\left(y+x\right)=\left(y-x\right).\left(-z\right)=\left(x-y\right).z\)
\(y=\left(z-y\right)\left(z+y\right)=\left(z-y\right).-x=x\left(y-z\right)\)
\(z=y\left(z-x\right)\)
=> \(xyz=\left(x-y\right)\left(y-z\right)\left(z-x\right).xyz\)
=> B = 1
\(\frac{x^3}{y}+xy\ge2\sqrt{\frac{x^3}{y}.xy}=2x^2\)
\(\Rightarrow\frac{x^3}{y}+\frac{y^3}{z}+\frac{z^3}{x}\ge2\left(x^2+y^2+z^2\right)-xy-yz-zx\ge2\left(x^2+y^2+z^2\right)-\left(xy+yz+zx\right)=1\)
Ta có: \(1+x^2=xy+yz+xz+x^2=\left(x+y\right)\left(x+z\right)\)
\(1+y^2=xy+yz+xz+y^2=\left(z+y\right)\left(x+y\right)\)
\(1+z^2=xy+yz+xz+z^2=\left(z+x\right)\left(z+y\right)\)
Thay vào biểu thức A, ta có bt sau:
\(A=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)
\(+y\sqrt{\frac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(y+z\right)\left(x+y\right)}}\)
\(+z\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(x+z\right)\left(z+y\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)(x,y,z dương)
\(=2\left(xy+xz+yz\right)=2.1=2\)
ban oi sai de ak vi x^2+y^2+z^2=3 roi thi can j tim max