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

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 \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

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

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

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 \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

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

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

Đặt \(\dfrac{y}{x}=a\ge4\)

\(Q=\dfrac{2a^2-2a+1}{a^2+a}=\dfrac{2a^2-2a+1}{a^2+a}-\dfrac{5}{4}+\dfrac{5}{4}=\dfrac{\left(a-4\right)\left(3a-1\right)}{4\left(a^2+1\right)}+\dfrac{5}{4}\ge\dfrac{5}{4}\)

\(Q_{min}=\dfrac{5}{4}\) khi \(a=4\) hay \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

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

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

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

\(A=\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)+\left(\dfrac{1}{2xy}+8xy\right)+\dfrac{3}{xy}\)

\(A\ge\dfrac{4}{x^2+y^2+2xy}+2\sqrt{\dfrac{8xy}{2xy}}+\dfrac{3}{\dfrac{1}{4}\left(x+y\right)^2}\ge20\)

\(A_{min}=20\) khi \(x=y=\dfrac{1}{2}\)

\(\dfrac{1}{x}+\dfrac{2}{y}\le1\Rightarrow\dfrac{2}{y}\le1-\dfrac{1}{x}\Rightarrow y\ge\dfrac{2x}{x-1}=2+\dfrac{2}{x-1}\)

\(x+\dfrac{2}{z}\le3\Rightarrow x< 3;\dfrac{2}{z}\le3-x\Rightarrow z\ge\dfrac{2}{3-x}\Rightarrow y+z\ge2+\dfrac{2}{x-1}+\dfrac{2}{3-x}\)

Lúc này ta sẽ áp dụng bất đẳng thức Bunhiacopski

Ta có:

\(6^2\le\left(y+z\right)^2=\left(\sqrt{2}\dfrac{y}{\sqrt{2}}Z\right)^2\le3\left(\dfrac{y^2}{2}+z^2\right)=\dfrac{3}{2}\left(y^2+2z^2\right)\)

\(\Rightarrow P\ge24\). Dấu đẳng thức xảy ra khi và chỉ khi \(y=4,z=2\) 

Vậy giá trị nhỏ nhật của P là 24

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

\(P=\dfrac{x+2y}{16}+\dfrac{1}{x+2y}+\dfrac{3\left(x+2y\right)}{16}\)

\(P\ge2\sqrt{\dfrac{x+2y}{16\left(x+2y\right)}}+\dfrac{3}{16}.2\sqrt{2xy}=\dfrac{5}{4}\)

\(P_{min}=\dfrac{5}{4}\) khi \(\left(x;y\right)=\left(2;1\right)\)