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$\frac{4a^2b^2}{(a^2+b^2)^2}+\frac{a^2}{b^2}+\frac{b^2}{a^2}$
$=\frac{4a^2b^2}{(a^2+b^2)^2}+\frac{1}{2}(\frac{a^2}{b^2}+\frac{b^2}{a^2})+\frac{1}{2}(\frac{a^2}{b^2}+\frac{b^2}{a^2})$
$=\frac{4a^2b^2}{(a^2+b^2)^2}+\frac{a^4+b^4}{2a^2b^2}+\frac{1}{2}(\frac{a^2}{b^2}+\frac{b^2}{a^2})$
$\geq \frac{4a^2b^2}{(a^2+b^2)^2}+\frac{(a^2+b^2)^2}{4a^2b^2}+1\geq 2+1=3$
Dấu bằng xảy ra khi $a=b$
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