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

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

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

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

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

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

Dấu "=" <=> x= y = 1/2

\(2,A=\frac{x^2+y^2}{xy}=\frac{x}{y}+\frac{y}{x}=\left(\frac{x}{9y}+\frac{y}{x}\right)+\frac{8x}{9y}\ge2\sqrt{\frac{x}{9y}.\frac{y}{x}}+\frac{8.3y}{9y}\)

                                                                                                  \(=2\sqrt{\frac{1}{9}}+\frac{8.3}{9}=\frac{10}{3}\)

Dấu "=" <=> x = 3y

bài 3 min hay max ?

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 \(x=y=\frac{1}{2}\)

Đề bài sai, biểu thức này ko có min

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

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