Áp dụng liên tục BĐT \(a^2+b^2\ge\frac{\left(a+b\right)^2}{2}\) kết hợp \(ab\le\frac{\left(a+b\right)^2}{4}\)
\(P=\left(x^4\right)^2+\left(y^4\right)^2+\frac{3}{xy}\ge\frac{\left(x^4+y^4\right)^2}{2}+\frac{3}{\frac{\left(x+y\right)^2}{4}}\ge\frac{\left(\frac{\left(x^2+y^2\right)^2}{2}\right)^2}{2}+\frac{12}{\left(x+y\right)^2}=\frac{\left(x^2+y^2\right)^4}{8}+3\)
\(P\ge\frac{\left(\frac{\left(x+y\right)^2}{2}\right)^4}{8}+3=\frac{\left(x+y\right)^8}{128}+3=5\)
\(\Rightarrow P_{min}=5\) khi \(x=y=1\)
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