Có: \(\left\{{}\begin{matrix}x^4+y^2\ge2x^{2y}\\x^2+y^4\ge2xy^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{x^4+y^2}\le\frac{x}{2x^{2y}}\\\frac{y}{x^2+y^4}\le\frac{y}{2xy^2}\end{matrix}\right.\)
Mà xy = 1 \(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2x^{2y}}=\frac{x}{2x}=\frac{1}{2}\\\frac{y}{2xy^2}=\frac{y}{2y}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\frac{x}{x^4+y^2}+\frac{y}{x^2+y^4}\le\frac{1}{2}+\frac{1}{2}=1\)
Vậy GTLN của A = 1
\("="\Leftrightarrow x=y=1\)
P/s: Bài này em không chắc chắn lắm, nhờ chị Akai Haruma kiểm tra giúp ạ.
\(xy=1\Rightarrow y=\frac{1}{x}\)
\(A=\frac{x}{x^4+\left(\frac{1}{x}\right)^2}+\frac{\frac{1}{x}}{x^2+\left(\frac{1}{x}\right)^4}=\frac{x^3}{x^6+1}+\frac{x^3}{x^6+1}=\frac{2x^3}{x^6+1}\le\frac{2x^3}{2\sqrt{x^6.1}}=\frac{2x^3}{2\left|x^3\right|}\le1\)
\(\Rightarrow A_{max}=1\) khi \(x=y=1\)
