Question

Cho x > 0; y > 0
Tìm GTNN của A = \(\frac{x^4}{y^4}+\frac{y^4}{x^4}-\frac{x^2}{y^2}-\frac{y^2}{x^2}+\frac{x}{y}+\frac{y}{x}\)

Answer 1

Đặt \(t=\frac{x}{y}+\frac{y}{x}\). Vì x; y > 0 => \(\frac{x}{y}>0;\frac{y}{x}>0\). Áp dung BDT Cô - si có:

\(t=\frac{x}{y}+\frac{y}{x}\ge2.\sqrt{\frac{x}{y}.\frac{y}{x}}=2\)

Có: \(\frac{x^2}{y^2}+\frac{y^2}{x^2}=\left(\frac{x}{y}+\frac{y}{x}\right)^2-2.\frac{x}{y}.\frac{y}{x}=t^2-2\)

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

Vậy \(A=t^4-4t^2+2-\left(t^2-2\right)+t=t^4-5t^2+t+4\)

=> \(A=\left(t^4-8t^2+16\right)+3t^2+t-12=\left(t^2-4\right)^2+3t^2+t-12=\left(t^2-4\right)^2+3\left(t^2-4\right)+t\ge2\)với mọi \(t\ge2\)

Vì \(t\ge2\) => \(t^2\ge4\Rightarrow t^2-4\ge0\)

Vậy Min A = 2 khi t = 2 <=> \(\frac{x}{y}+\frac{y}{x}=2\) <=> x = y = 1