Question

[Phạm Dương Ngọc Nhi]

Cho x,y,z > 0, x + y + z \(\ge\)1 . Chứng minh :

\(\frac{x^5}{y^4}+\frac{y^5}{z^4}+\frac{z^5}{x^4}\ge1\)

Answer 1

[Nguyễn Việt Lâm]

\(\frac{x^5}{y^4}+\frac{x^5}{y^4}+y+y+y\ge5\sqrt[5]{\frac{x^{10}y^3}{y^8}}=\frac{5x^2}{y}\)

Tương tự: \(\frac{2y^5}{z^4}+3z\ge\frac{5y^2}{z}\) ; \(\frac{2z^5}{x^4}+3x\ge\frac{5z^2}{x}\)

Cộng vế với vế:

\(2\left(\frac{x^5}{y^4}+\frac{y^5}{z^4}+\frac{z^5}{x^4}\right)+3\left(x+y+z\right)\ge5\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\right)\ge5\left(x+y+z\right)\)

\(\Rightarrow2\left(\frac{x^5}{y^4}+\frac{y^5}{z^4}+\frac{z^5}{x^4}\right)\ge2\left(x+y+z\right)\ge2\)

\(\Rightarrow\frac{x^5}{y^4}+\frac{y^5}{z^4}+\frac{z^5}{x^4}\ge1\)

Dấu "=" xay ra khi \(x=y=z=\frac{1}{3}\)