\(P=\dfrac{a^2b^2}{a^2+b^2+a^4b^2+a^2b^4}=\dfrac{a^2b^2}{a^2\left(a^2b^2+1\right)+b^2\left(a^2b^2+1\right)}=\dfrac{a^2b^2}{\left(a^2+b^2\right)\left(a^2b^2+1\right)}\)-Áp dụng BĐT AM-GM ta có:
\(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\a^2b^2+1\ge2ab\end{matrix}\right.\)
-Do đó: \(P=\dfrac{a^2b^2}{\left(a^2+b^2\right)\left(a^2b^2+1\right)}\le\dfrac{a^2b^2}{2ab.2ab}=\dfrac{1}{4}\)
\(P=\dfrac{1}{4}\Leftrightarrow a=b=1\)
-Vậy \(P_{max}=\dfrac{1}{4}\)
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