\(K=\left(4xy+\dfrac{1}{4xy}\right)+\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\dfrac{5}{4xy}\)

\(K\ge2\sqrt{\dfrac{4xy}{4xy}}+\dfrac{4}{x^2+y^2+2xy}+\dfrac{5}{\left(x+y\right)^2}\ge2+4+5=11\)

\(K_{min}=11\) khi \(x=y=\dfrac{1}{2}\)

\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le\dfrac{1}{4}\)

Đặt \(\dfrac{x}{y}=a\Rightarrow0< a\le\dfrac{1}{4}\)

\(P=\dfrac{\left(\dfrac{x}{y}\right)^2-\dfrac{2x}{y}+2}{\dfrac{x}{y}+1}=\dfrac{a^2-2a+2}{a+1}=\dfrac{4a^2-8a+8}{4\left(a+1\right)}=\dfrac{4a^2-13a+3+5\left(a+1\right)}{4\left(a+1\right)}\)

\(P=\dfrac{5}{4}+\dfrac{\left(1-4a\right)\left(3-a\right)}{4\left(a+1\right)}\ge\dfrac{5}{4}\)

Dấu "=" xảy ra khi \(a=\dfrac{1}{4}\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

Áp dụng BĐT cosi:

`1/x^2+1/y^2>=2/(xy)`

`<=>2>=2/(xy)`

`<=>1>=1/(xy)`

`<=>xy>=1`

Dấu "=" xảy ra khi `x=y=1`

\(1\ge x+\dfrac{1}{y}\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le\dfrac{1}{4}\Rightarrow\dfrac{y}{x}\ge4\)

\(P=\dfrac{1-\dfrac{2y}{x}+2\left(\dfrac{y}{x}\right)^2}{1+\dfrac{y}{x}}\)

Đặt \(\dfrac{y}{x}=a\ge4\Rightarrow P=\dfrac{2a^2-2a+1}{a+1}=2a-4+\dfrac{5}{a+1}\)

\(P=\dfrac{a+1}{5}+\dfrac{5}{a+1}+\dfrac{9}{5}.a-\dfrac{21}{5}\ge2\sqrt{\dfrac{5\left(a+1\right)}{5\left(a+1\right)}}+\dfrac{9}{5}.4-\dfrac{21}{5}=5\)

Dấu "=" xảy ra khi \(a=4\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

Áp dụng cosi

`1/x^2+1/y^2>=2/(xy)`

`=>1/2>=2/(xy)`

`=>xy>=4`

Aps dụng cosi

`=>x+y>=2\sqrt{xy}=2.2=4`

Dấu "=" xảy ra khi `x=y=4`

Có : \(\dfrac{1}{2}=\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge2\sqrt{\dfrac{1}{x^2}\cdot\dfrac{1}{y^2}}=\dfrac{2}{xy}\)

\(\Rightarrow xy\ge4\)

Ta có : \(A=x+y\ge2\sqrt{xy}=2\sqrt{4}=4\)

Dấu "=" xảy ra khi \(x=y=2\)

Vậy min A = 4 khi $x=y=2$

\(P=\frac{1}{x^2+y^2}+\frac{2}{xy}+4xy\)

\(=\frac{1}{x^2+y^2}+\frac{1}{2xy}+4xy+\frac{3}{2xy}\)

\(\ge\frac{4}{x^2+y^2+2xy}+4xy+\frac{1}{4xy}+\frac{5}{4xy}\)

\(\ge\frac{4}{x^2+y^2+2xy}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{5}{4xy}\)

Ta có BĐT phụ: \(\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\)(đúng )

Dấu "=" xảy ra <=> x=y

\(\Rightarrow P\ge\frac{4}{\left(x+y\right)^2}+2+\frac{5}{\left(x+y\right)^2}\)

\(\ge\frac{4}{1}+2+\frac{5}{1}=11\)

Dấu"=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)

Vậy Min P =11 \(\Leftrightarrow x=y=\frac{1}{2}\)