Question

Tìm x, y, z biết rằng: \(x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=4\)

Answer 1

ĐK: x,y khác 0

Áp dụng BĐT Cô-si ta có:

\(x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}\\ \ge2\sqrt{x^2.\dfrac{1}{x^2}}+2\sqrt{y^2.\dfrac{1}{y^2}}\\ =2+2=4\)

Dấu bằng xảy ra khi và chỉ khi: \(x=y=\pm1\)

Answer 2

Ta có:

\(x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=4\\ \Leftrightarrow x^2-2+\dfrac{1}{x^2}+y^2-2+\dfrac{1}{y^2}=0\\ \Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)

Do \(\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\) và \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{x}\right)^2\ge0\\\left(y-\dfrac{1}{y}\right)^2\ge0\end{matrix}\right.\) nên:

\(\left(x-\dfrac{1}{x}\right)^2=\left(y-\dfrac{1}{y}\right)^2=0\)

Do đó: \(x=y=\pm1\)