\(P=\dfrac{x^2}{1+x^4}\)
Áp dụng BDT Cô-si: \(a^2+b^2\ge2ab\)
\(\Rightarrow x^4+1\ge2x^2\\ \Rightarrow P\le\dfrac{x^2}{2x^2}\le\dfrac{1}{2}\)
Dấu "=" xảy ra khi:
\(x^4=1\\ \Leftrightarrow x=\pm1\)
Vậy \(P_{Max}=\dfrac{1}{2}\) khi \(x=\pm1\)