Question

[Nguyễn Quỳnh Anh]

giúp mình với ạ 

Answer 1

[Đoàn Đức Hà]

\(Q=\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}\)

Ta có: 

\(\frac{x^4}{\left(x^2+y^2\right)\left(x+y\right)}-\frac{y^4}{\left(x^2+y^2\right)\left(x+y\right)}=\frac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}=x-y\)

Tương tự: 

\(\frac{y^4}{\left(y^2+z^2\right)\left(y+z\right)}-\frac{z^4}{\left(y^2+z^2\right)\left(y+z\right)}=y-z\)

\(\frac{z^4}{\left(z^2+x^2\right)\left(z+x\right)}-\frac{x^4}{\left(z^2+x^2\right)\left(z+x\right)}=z-x\)

Cộng lại vế với vế ta được: 

\(\frac{x^4-y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4-z^4}{\left(y^2+z^2\right)\left(y-z\right)}+\frac{z^4-x^4}{\left(z^2+x^2\right)\left(z+x\right)}=0\)

Suy ra \(2Q=\frac{x^4+y^4}{\left(x^2+y^2\right)\left(x+y\right)}+\frac{y^4+z^4}{\left(y^2+z^2\right)\left(y+z\right)}+\frac{z^4+x^4}{\left(z^2+x^2\right)\left(z+x\right)}\)

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

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

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

\(=\frac{x+y}{4}+\frac{y+z}{4}+\frac{z+x}{4}\)

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

Dấu \(=\)xảy ra khi \(x=y=z=\frac{1}{3}\).

Vậy \(minQ=\frac{1}{4}\)đạt tại \(x=y=z=\frac{1}{3}\).