cách 4$\frac{a^{2}}{b^{2}}+\frac{a^{2}}{b^{2}}+\frac{b}{a}\ge3\sqrt[3]{\frac{a^{2}}{b^{2}}\frac{a^{2}}{b^{2}}\frac{b}{a}}=3\frac{a}b$
$\frac{a^{2}}{a^{2}}+\frac{b^{2}}{a^{2}}+\frac{a}{b}\ge3\sqrt[3]{\frac{b^{2}}{a^{2}}\frac{b^{2}}{a^{2}}\frac{a}{b}}=3\frac{b}a$
$\Rightarrow 2(\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}})+\frac{a}{b}+\frac{b}{a}\ge 3(\frac{a}{b}+\frac{b}{a})$
$\Leftrightarrow\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}\geq \frac{a}{b}+\frac{b}{a}$