Question

Cho x,y,z la các số dương sao cho x+y+z\(\ge\)12

tìm Min M=\(\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{z}}+\frac{z}{\sqrt{x}}\)

Answer 1

\(P=4\left(\frac{x}{y+4}+\frac{y}{z+4}+\frac{z}{x+4}\right)=4\left(\frac{x^2}{xy+4x}+\frac{y^2}{yz+4y}+\frac{z^2}{zx+4z}\right)\)

\(\ge\frac{4\left(a+b+c\right)^2}{xy+4x+yz+4y+zx+4z}=\frac{4.12^2}{4.12+\left(xy+yz+zx\right)}\)

\(\ge\frac{4.12^2}{4.12+\frac{\left(x+y+z\right)^2}{3}}=\frac{4.12^2}{4.12+\frac{12^2}{3}}=6\)

Answer 2

Ta có

\(\frac{x}{\sqrt{y}}+\frac{x}{\sqrt{y}}+\frac{xy}{8}\ge3\sqrt[3]{\frac{x}{\sqrt{y}}.\frac{x}{\sqrt{y}}.\frac{xy}{8}}=\frac{3x}{2}\)

Tương tự cho 2 cái kia

Cộng lại theo vế:

\(2M\ge\frac{3}{2}\left(x+y+z\right)-\frac{xy+yz+zx}{8}\ge\frac{3}{2}\left(x+y+z\right)-\frac{\left(x+y+z\right)^2}{24}\ge12\)

Vậy  \(M\ge6\)

Answer 3

Giải lại

Ta có

\(M^2=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+2\left(\frac{xy}{\sqrt{yz}}+\frac{yz}{\sqrt{zx}}+\frac{zx}{\sqrt{xy}}\right)\)

Lại có

\(\hept{\begin{cases}\frac{xy}{\sqrt{yz}}+\sqrt{yz}\ge2\sqrt{xy}\\\frac{yz}{\sqrt{zx}}+\sqrt{zx}\ge2\sqrt{yz}\\\frac{zx}{\sqrt{xy}}+\sqrt{xy}\ge2\sqrt{zx}\end{cases}}\)

Cộng theo vế suy ra  \(\frac{xy}{\sqrt{yz}}+\frac{yz}{\sqrt{zx}}+\frac{zx}{\sqrt{xy}}\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

Do đó

\(M^2=\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+2\left(\frac{xy}{\sqrt{yz}}+\frac{yz}{\sqrt{zx}}+\frac{zx}{\sqrt{xy}}\right)\)

\(\ge\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)

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

\(\ge3\sqrt[3]{\frac{x^2}{y}.\sqrt{xy}.\sqrt{xy}}+3\sqrt[3]{\frac{y^2}{z}.\sqrt{yz}.\sqrt{yz}}+3\sqrt[3]{\frac{z^2}{x}.\sqrt{zx}.\sqrt{zx}}\)

\(=3\left(x+y+z\right)\ge36\)

Vậy  \(M\ge6\)

ĐT xảy ra tại  \(x=y=z=4\)

Answer 4

Cách khác :D

(continue cách đầu tiên)

\(P\ge\frac{4\left(x+y+z\right)^2}{\left(xy+yz+zx\right)+4\left(x+y+z\right)}\ge\frac{4\left(x+y+z\right)^2}{\frac{\left(x+y+z\right)^2}{3}+4\left(x+y+z\right)}=\frac{4\left(x+y+z\right)}{\frac{x+y+z}{3}+4}\)

\(=\frac{4\left(x+y+z\right)}{\frac{x+y+z+12}{3}}=\frac{12\left(x+y+z\right)}{\left(x+y+z\right)+12}\ge\frac{12\left(x+y+z\right)}{\left(x+y+z\right)+\left(x+y+z\right)}=6\)