\(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\ge2\)
Xét hiệu:
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}-2=\dfrac{a^4}{a^2b^2}+\dfrac{b^4}{a^2b^2}-\dfrac{2a^2b^2}{a^2b^2}\)
\(=\dfrac{a^4+b^4-2a^2b^2}{a^2b^2}=\dfrac{\left(a^4-2a^2b^2+b^4\right)}{a^2b^2}\)
\(=\dfrac{\left(a^2-b^2\right)^2}{\left(ab\right)^2}=\left(\dfrac{a^2-b^2}{ab}\right)^2\ge0\)
=> BĐT luôn đúng