\(x+y+2xy=\dfrac{15}{2}\)\(\Rightarrow\dfrac{15}{2}\le\left(x+y\right)+\dfrac{\left(x+y\right)^2}{2}\)

\(\Leftrightarrow\left(x+y\right)^2+2\left(x+y\right)-15\ge0\)

\(\Leftrightarrow\left(x+y+5\right)\left(x+y-3\right)\ge0\)

\(\Leftrightarrow x+y\ge3\) (vì \(x+y+5>0\) với mọi x,y dương)

\(\Rightarrow P_{min}=3\)

Dấu = xảy ra <=> \(x=y=\dfrac{3}{2}\)

Tìm GTLN:

Xét hiệu $2.(x^2+y^2)-(x+y)^2=2.(x^2+y^2)-x^2-y^2-2xy=x^2-2xy+y^2=(x-y)^2 \geq 0$

Nên $(x+y)^2 \leq 2.(x^2+y^2)=2$ (do $x^2+y^2=1$)

Dấu $=$ xảy ra $⇔(x-y)^2=0;x^2+y^2=1⇔x=y;x^2+y^2=1⇔x=y=\dfrac{1}{\sqrt[]2}$

Tìm Min:

Có $(x+y)^2 \geq 0$ với mọi $x;y$

Dấu $=$ xảy ra $⇔(x+y)^2=0;x^2+y^2=0⇔x=-y;x^2+y^2=1⇔x=\dfrac{1}{\sqrt[]2};y=-\dfrac{1}{\sqrt[]2}$ và hoán vị

\(P=\dfrac{1}{2023}\dfrac{1}{z}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=\dfrac{1}{2023.z}\dfrac{x+y}{xy}\)

Ap dung BDT cosi taco 

\(P\ge\dfrac{1}{2023z}.\dfrac{x+y}{\dfrac{\left(x+y\right)^2}{4}}=\dfrac{4}{2023z}\dfrac{1}{x+y}\)

<->\(P\ge\dfrac{4}{2023}\dfrac{1}{z\left(1-z\right)}=\dfrac{4}{2023}\dfrac{1}{-z^2+z}=\dfrac{4}{2023}\dfrac{1}{-\left(z-\dfrac{1}{2}\right)^2+\dfrac{1}{4}}\)

\(< =>P\ge\dfrac{4}{2023}\dfrac{1}{\dfrac{1}{4}}=\dfrac{16}{2023}\)

\(P_{min}=\dfrac{16}{2023}\Leftrightarrow Z=\dfrac{1}{2},x=y=\dfrac{1}{4}\)

\(P=\dfrac{6x+6y+2xy}{2}=\dfrac{6x+6y+2xy+10-10}{2}\)

\(=\dfrac{6x+6y+2xy+2\left(x^2+y^2\right)+6}{2}-5\)

\(=\dfrac{\left(x+y+2\right)^2+\left(x+1\right)^2+\left(y+1\right)^2}{2}-5\ge-5\)

\(P_{min}=-5\) khi \(x=y=-1\)

BẠN THAM KHẢO :

Đáp án D

Theo bất đẳng thức Cauchy – Schwarz có

Do đó