Với a;b > 0 ; AD BĐT Cô-si ; ta được : \(a^4+1+1+1\ge4a\) \(\Rightarrow a^4+3\ge4a\)
\(\Rightarrow\dfrac{1}{a^4+4}\le\dfrac{1}{4a+1}\) . CMTT : \(\dfrac{1}{b^4+4}\le\dfrac{1}{4b+1}\)
Suy ra : \(A=\dfrac{1}{a^4+4}+\dfrac{1}{b^4+4}\le\dfrac{1}{4a+1}+\dfrac{1}{4b+1}\)
Mặt khác : \(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+1\ge\dfrac{\left(1+1+1+1+1\right)^2}{4a+1}\) ( B.C.S)
\(\Rightarrow\dfrac{4}{a}+1\ge\dfrac{25}{4a+1}\Rightarrow\dfrac{1}{25}\left(\dfrac{4}{a}+1\right)\ge\dfrac{1}{4a+1}\)
CMTT : \(\dfrac{1}{4b+1}\le\dfrac{1}{25}\left(\dfrac{4}{b}+1\right)\)
Suy ra : \(A\le\dfrac{4}{25}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{2}{25}=\dfrac{4}{25}.2+\dfrac{2}{25}=\dfrac{2}{5}\)
" = " \(\Leftrightarrow a=b=1\)