3: 

Ta có: \(\left(2x+1\right)^2\ge0\forall x\)

\(\Leftrightarrow\left(2x+1\right)^2+2021\ge2021\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

Ta có: \(\left(x-1\right)^2+\left(x+y\right)^2\le9\Rightarrow x+y\le3\).

Áp dụng bất đẳng thức AM - GM ta có:

\(\dfrac{2}{x}+2x\ge2\sqrt{\dfrac{2}{x}.2x}=4;\dfrac{4}{y}+y\ge2\sqrt{\dfrac{4}{y}.y}=4\).

Do đó \(\dfrac{2}{x}\ge4-2x;\dfrac{4}{y}\ge4-y\)

\(\Rightarrow P\ge8-4\left(x+y\right)\ge-4\). (do \(x+y\le3\)).

Vậy...

Đẳng thức xảy ra khi và chỉ khi x = 1; y = 2.

\(2x^2+2x-2xy+y^4+4=x^2+2x+1+x^2-2xy+y^2+y^4-2\cdot y^2\cdot\frac{1}{2}+\frac{1}{4}+\frac{11}{8}\)

\(\sqrt{2}A=\sqrt{4x^2-4x+10}+\sqrt{4x^2-8x+8}\)

\(\sqrt{2}A=\sqrt{\left(2x-1\right)^2+3^2}+\sqrt{\left(2-2x\right)^2+2^2}\)

Áp dụng BĐT \(\sqrt{A^2+B^2}+\sqrt{C^2+D^2}\ge\sqrt{\left(A+C\right)^2+\left(B+D\right)^2}\)

=>\(\sqrt{2}A\ge\sqrt{\left(2x-1+2-2x\right)^2+\left(3+2\right)^2}=\sqrt{26}\)

=>\(A\ge\sqrt{13}\)

Dấu bằng xảy ra<=> \(\frac{2x-1}{3}=\frac{2x-2}{2}\)

<=>.........

12x^2(2x - y) - 4xy(2x + y) = 4x(x- y)(y + 6x)

\(M=x^2+4x+4+y^2-4y=\left(2x^2+4x+2\right)+\left(y^2-4y+4\right)-2\)

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