Áp dụng BĐT Bunhiacopski ta có:
\(\sqrt{x^2+\frac{1}{x^2}}=\frac{1}{\sqrt{17}}\sqrt{\left(x^2+\frac{1}{x^2}\right)\left(4^2+1^2\right)}\ge\frac{1}{\sqrt{17}}\left(4x+\frac{1}{x}\right)\)
Tương tự:
\(\sqrt{y^2+\frac{1}{y^2}}\ge\frac{1}{\sqrt{17}}\left(4y+\frac{1}{y}\right)\)
Cộng lại ta được:
\(\sqrt{x^2+\frac{1}{x^2}}+\sqrt{y^2+\frac{1}{y^2}}\ge\frac{1}{\sqrt{17}}\left(4x+4y+\frac{1}{x}+\frac{1}{y}\right)\)
\(\ge\frac{1}{\sqrt{17}}\left[4\left(x+y\right)+\frac{4}{x+y}\right]=\frac{1}{\sqrt{17}}\left(16+1\right)=\sqrt{17}\)
Dấu "=" xảy ra tại x=y=2
