Question

Tìm giá trị nhỏ nhất của biểu thức \(P=\frac{x^2}{y^2}+\frac{y^2}{x^2}-3\left(\frac{x}{y}+\frac{y}{x}\right)+5\) (với \(x\ne0,y\ne0\))

Answer 1

\(P=\frac{x^2}{y^2}+\frac{y^2}{x^2}-3\left(\frac{x}{y}+\frac{y}{x}\right)+5\)

\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)^2-2-3\left(\frac{x}{y}+\frac{y}{x}\right)+5\)

\(\Leftrightarrow\left(\frac{x}{y}+\frac{y}{x}\right)\left(\frac{x}{y}+\frac{y}{x}-3\right)+3\)

Ta có: \(\left(\frac{x}{y}+\frac{y}{x}\right)\ge2\Rightarrow\left(\frac{x}{y}+\frac{y}{x}-3\right)\ge-1\Rightarrow\left(\frac{x}{y}+\frac{y}{x}\right)\left(\frac{x}{y}+\frac{y}{x}-3\right)\ge-2\)

\(\Rightarrow\left(\frac{x}{y}+\frac{y}{x}\right)\left(\frac{x}{y}+\frac{y}{x}-3\right)+3\ge1\)

\(\Rightarrow P\ge1\)

Vậy \(Min_P=1\)

Answer 2

\(ĐK:x,y>0\)

Answer 3

Áp dụng bất đẳng thức cosi cho 2 số dương ta có :

P>=\(2\sqrt{\frac{x^2\cdot y^2}{y^2\cdot x^2}}-3\cdot2\cdot\sqrt{\frac{x\cdot y}{y\cdot x}}+5=2-6+5=1\)

Vậy Min P =1 . dấu = xảy ra khi x=y=1

Answer 4

Thêm đk: x;y >0

\(P=\)\(\frac{x^2}{y^2}+\frac{y^2}{x^2}-3\left(\frac{x}{y}+\frac{y}{x}\right)+5\)

\(=\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}+2\right)-3\left(\frac{x}{y}+\frac{y}{x}\right)+3\)

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

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

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

Ta có: \(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}=2\)

\(\Rightarrow P\ge1\)

Dấu " = " xảy ra <=> x=y