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

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

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 ?

A= \(\frac{1}{\left(x+y\right)\left(x^2+y^2-xy\right)+xy}+\frac{4x^2y^2+2}{xy}=\)\(\frac{1}{x^2+y^2}+4xy+\frac{2}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+4xy+\frac{1}{4xy}+\frac{5}{4xy}\) (1)

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b};a+b\ge2\sqrt{ab},\frac{1}{xy}\ge\frac{4}{\left(x+y\right)^2}\)áp dụng vào trên ta được

 (1) \(\ge\frac{4}{x^2+y^2+2xy}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{5}{4}.\frac{4}{\left(x+y\right)^2}=4+2+\frac{5}{4}.4=11.\)

dấu '=" khi x=y = 1/2

#)Giải :

Ta có : \(x^2+y^2-xy=4\Leftrightarrow x^2+y^2=4+xy\Leftrightarrow3\left(x^2+y^2\right)=8\left(x+y\right)^2\ge8\)

\(\Rightarrow A_{max}=8\)

Dấu''='' xảy ra khi x = y = 2 hoặc x = y = -2

\(=>x^2+y^2-xy=4=x^2+y^2=4+xy=3\left(x^2+y^2\right)=8\left(x+y\right)^2>8\)

\(=>A=8\)

~Study well~ :)

\(A=x^2+xy+y^2-3(x+y)+3\\2A=2x^2+2xy+2y^2-6(x+y)+6\\=(x^2+2xy+y^2)-4(x+y)+4+(x^2-2x+1)+(y^2-2y+1)\\=(x+y)^2-4(x+y)+4+(x-1)^2+(y-1)^2\\=(x+y-2)^2+(x-1)^2+(y-1)^2\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x+y-2\right)^2\ge0\forall x,y\\\left(x-1\right)^2\ge0\forall x\\\left(y-1\right)^2\ge0\forall y\end{matrix}\right.\)

\(\Rightarrow\left(x+y-2\right)^2+\left(x-1\right)^2+\left(y-1\right)^2\ge0\forall x,y\)

\(\Rightarrow2A\ge0\forall x,y\)

\(\Rightarrow A\ge0\forall x,y\)

Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}x+y-2=0\\x-1=0\\y-1=0\end{matrix}\right.\Rightarrow x=y=1\)

Vậy \(Min_A=0\) khi \(x=y=1\).

\(\text{#}Toru\)

\(2A=2x^2+2y^2+2xy-6x-6y+6\)

\(2A=\left(x+y\right)^2-4\left(x+y\right)+4+\left(x-1\right)^2+\left(y-1\right)^2\)

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

Do \(\left\{{}\begin{matrix}\left(x+y-2\right)^2\ge0\\\left(x-1\right)^2\ge0\\\left(y-1\right)^2\ge0\end{matrix}\right.\) ;\(\forall x;y\)

\(\Rightarrow2A\ge0\Rightarrow A\ge0\)

Vậy \(A_{min}=0\) khi \(\left\{{}\begin{matrix}x+y-2=0\\x-1=0\\y-1=0\end{matrix}\right.\) hay \(\left(x;y\right)=\left(1;1\right)\)

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

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

\(=3+\dfrac{2xy}{x^2+y^2}+\dfrac{x^2+y^2}{2xy}+\dfrac{x^2+y^2}{2xy}\)

\(\dfrac{2xy}{x^2+y^2}+\dfrac{x^2+y^2}{2xy}>=2\cdot\sqrt{\dfrac{2xy}{x^2+y^2}\cdot\dfrac{x^2+y^2}{2xy}}=2\)

Dấu = xảy ra khi \(\dfrac{x^2+y^2}{2xy}=\dfrac{2xy}{x^2+y^2}\)

=>x=y

x^2+y^2>=2xy

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

Dấu = xảy ra khi x=y

=>S>=6

Dấu = xảy ra khi x=y

usechatgpt init success là gì vậy bạn :))?

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

\(\Rightarrow P=8-\left(x-y\right)^2\le8\)

\(MaxP=8\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-xy=4\\x-y=0\end{matrix}\right.\Leftrightarrow x=y=\pm2\)

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

\(\Rightarrow4=\dfrac{3}{2}P-\dfrac{1}{2}\left(x+y\right)^2\)

\(\Rightarrow P=\dfrac{8+\left(x+y\right)^2}{3}\ge\dfrac{8}{3}\)

\(MinP=\dfrac{8}{3}\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2-xy=4\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\pm\dfrac{2\sqrt{3}}{3}\\y=\mp\dfrac{2\sqrt{3}}{3}\end{matrix}\right.\)

Lời giải:

Tìm min:
Áp dụng BĐT AM-GM:
$x^2+y^2=4+xy\leq 4+|xy|\leq 4+\frac{x^2+y^2}{2}$

$\Rightarrow \frac{x^2+y^2}{2}\leq 4$

$\Rightarrow P=x^2+y^2\leq 8$

Vậy $P_{\max}=8$

---------------------------

$P=x^2+y^2=\frac{2}{3}(x^2-xy+y^2)+\frac{1}{3}(x^2+2xy+y^2)$

$=\frac{2}{3}.4+\frac{1}{3}(x+y)^2=\frac{8}{3}+\frac{1}{3}(x+y)^2\geq \frac{8}{3}$
Vậy $P_{\min}=\frac{8}{3}$