A=2x^2+(y+1)^4+1
Ta thấy:\(\begin{cases}2x^2\\\left(y+1\right)^4\end{cases}\ge0\)
\(\Rightarrow2x^2+\left(y+1\right)^4\ge0+0=0\)
\(\Rightarrow2x^2+\left(y+1\right)^4+1\ge0+1=1\)
\(\Rightarrow A\ge1\)
Dấu = khi \(\begin{cases}2x^2=0\\\left(y+1\right)^4=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=0\\y=-1\end{cases}\)
Vậy MinA=1 khi x=0; y=-1