Ta có :

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

Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}>=\frac{4}{a+b}\)( "=" khi a=b ) , ta có :

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

\(\Rightarrow\frac{1}{x^2+y^2}+\frac{1}{2xy}>=\frac{4}{\left(x+y\right)^2}=\frac{4}{1^2}=4\)    (2)  

Lại có : \(\left(x-y\right)^2>=0\) ("=" khi x=y )

\(\Leftrightarrow x^2-2xy+y^2>=0\)

\(\Leftrightarrow x^2+y^2>=2xy\)

\(\Leftrightarrow x^2+y^2+2xy>=4xy\)

\(\Leftrightarrow\left(x+y\right)^2>=4xy\)

\(\Leftrightarrow1>=4xy\)

\(\Leftrightarrow2xy< =\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{2xy}>=2\)  (3)

Từ (1) , (2) và (3) , suy ra :  \(K>=4+2=6\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x^2+y^2=2xy\\x=y\\x+y=1\end{cases}}\)

                             \(\Rightarrow x=y=\frac{1}{2}\)

        Vậy Min\(K=6\)khi \(x=y=\frac{1}{2}\)

\(P=\dfrac{1}{x^2+y^2+z^2}+\dfrac{2023}{xy+yz+zx}\)

\(=\dfrac{1}{x^2+y^2+z^2}+\dfrac{1}{xy+yz+zx}+\dfrac{1}{xy+yz+zx}+\dfrac{2021}{xy+yz+zx}\)

\(\ge\dfrac{9}{\left(x+y+z\right)^2}+\dfrac{2021}{\dfrac{\left(x+y+z\right)^2}{3}}\)\(=9+\dfrac{2021}{\dfrac{1}{3}}=6072\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{3}\)

Ta có:

+) \(xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}\left(\text{Cô si}\right)\)

+) \(\dfrac{1}{x^2+y^2+z^2}+\dfrac{1}{xy+yz+zx}+\dfrac{1}{xy+yz+zx}\)

\(\ge\dfrac{9}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}=\dfrac{9}{\left(x+y+z\right)^2}\left(\text{Svácxơ}\right)\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel có:

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\ge\dfrac{4}{x^2+y^2+2xy}+\dfrac{1}{\dfrac{\left(x+y\right)^2}{2}}=\dfrac{4}{\left(x+y\right)^2}+\dfrac{2}{\left(x+y\right)^2}=6\)

Dấu "=" xảy ra khi x=y=\(\dfrac{1}{2}\)

áp dụng BDT AM-GM

\(=>x+y\ge2\sqrt{xy}=>1\ge2\sqrt{xy}=>\sqrt{xy}\le\dfrac{1}{2}=>xy\le\dfrac{1}{4}\)

\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\)

\(\ge\dfrac{4}{x^2+2xy+y^2}+\dfrac{1}{2.\dfrac{1}{4}}=\dfrac{4}{\left(x+y\right)^2}+2=4+2=6\)

dấu"=" xảy ra \(< =>x=y=\dfrac{1}{2}\)

h

\(A=\dfrac{\left(x-y\right)^2+2xy}{x-y}=x-y+\dfrac{2xy}{x-y}=x-y+\dfrac{2}{x-y}>=2\sqrt{2}\)

Dấu = xảy ra khi \(\left\{{}\begin{matrix}x=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\y=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)

Đề bài sai, biểu thức này ko có min

\(x+y=xy\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}=1\)

Đặt \(\left(\dfrac{1}{x};\dfrac{1}{y}\right)=\left(a;b\right)\Rightarrow a+b=1\) \(\Rightarrow a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2=\dfrac{1}{2}\)

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

Dấu "=" xảy ra khi \(x=y=2\)

\(P=\sum\dfrac{1}{x+y+1}\ge\dfrac{9}{2\left(x+y+z\right)+3}=\dfrac{9}{2.1+3}=\dfrac{9}{5}\)

Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)

Lời giải:

Áp dụng BĐT Cô-si:

\(x^2+y^2+z^2\geq \frac{(x+y+z)^2}{3}\)

\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\geq \frac{1}{3}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})^2\geq \frac{1}{3}.(\frac{9}{x+y+z})^2=\frac{27}{(x+y+z)^2}\)

\(\Rightarrow P\geq \frac{(x+y+z)^2}{3}+\frac{27}{(x+y+z)^2}\)

Áp dụng BĐT Cô-si:

\(\frac{(x+y+z)^2}{3}+\frac{1}{3(x+y+z)^2}\geq \frac{2}{3}\)

\(\frac{80}{3(x+y+z)^2}\geq \frac{80}{3}\)

\(\Rightarrow P\geq \frac{2}{3}+\frac{80}{3}=\frac{82}{3}\)

Vậy $P_{\min}=\frac{82}{3}$ khi $x=y=z=\frac{1}{3}$