Áp dụng Bunyakovsky, ta có :

\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)

=> \(\left(x^2+y^2\right)\ge\frac{1}{2}\)

=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)

Mấy cái kia tương tự 

\(F=\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+2021\\ F=\left(x-y\right)^2+\left(y-1\right)^2+2021\ge2021\)

Dấu \("="\Leftrightarrow x=y=1\)

Vậy \(F_{min}=2021\)

\(\Rightarrow F=\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+2021\\ \Rightarrow F=\left(x-y\right)^2+\left(y-1\right)^2+2021\ge2021\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y\\y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

((X+1)^2)^2 bé hơn hoặc bằng 0

Suy ra x+1=0,Nên x=-1

gọi biểu thức trên là A.

Ta có: \(A=x^2-2xy+2y^2-6y+9\)

\(\Rightarrow A=x^2-2xy+y^2+y^2-6y+9\)

\(\Rightarrow A=\left(x^2-2xy+y^2\right)+\left(y^2-6y+9\right)\)

 \(A=\left(x-y\right)^2+\left(y-3\right)^2\)

Nhận xét: \(\left(x+y\right)^2\ge0\forall x,y\)

                 \(\left(y-3\right)^2\ge0\forall y\)

\(\Rightarrow\left(x+y\right)^2+\left(y-3\right)^2\ge0\forall x,y\)

Vậy \(minA=0\) khi \(y-3=0\Rightarrow y=3\)

                                       \(x-y=0\Rightarrow x-3=0\Rightarrow x=3\)

KL: Vậy \(minA=0\) khi \(x=3;y=3\)

Đặt \(A=x^2-2xy+2y^2-6y+9=\left(x^2-2xy+y^2\right)+\left(y^2-6y+9\right)=\left(x-y\right)^2+\left(y-3\right)^2\)

Vì \(\left(x-y\right)^2\ge0;\left(y-3\right)^2\ge0\Rightarrow A=\left(x-y\right)^2+\left(y-3\right)^2\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-y=0\\y-3=0\end{cases}\Leftrightarrow x=y=3}\)

Vậy Amin = 0 khi x = y = 3