Áp dụng bđt Bunhiacopxki , ta có : \(4=\left(x+y\right)^2=\left(\frac{1}{\sqrt{3}}.\sqrt{3}.x+1.y\right)^2\le\left[\left(\frac{1}{\sqrt{3}}\right)^2+1^2\right].\left(3x^2+y^2\right)\)

\(\Rightarrow3x^2+y^2\ge\frac{4}{\frac{1}{3}+1}=3\) \(\Rightarrow A\ge3\)

Vậy Min A = 3 \(\Leftrightarrow\hept{\begin{cases}x+y=2\\3x=y\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{3}{2}\end{cases}}\)

x+y=2=>y=2-x

A=3x²+y²

A=3x²+(2-x)²

A=4x²-4x+4

A=(2x-1)²+3

Vậy: Min_A=3

3 nhé

đúng rùi đó Nguyễn Văn Tân

rồi đó

\(A=\dfrac{3x^2-2xy}{x^2+2xy+y^2}=\dfrac{15x^2-10xy}{5\left(x^2+2xy+y^2\right)}=\dfrac{-\left(x^2+2xy+y^2\right)+16x^2-8xy+y^2}{5\left(x^2+2xy+y^2\right)}\)

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

\(A_{min}=-\dfrac{1}{5}\) khi \(4x-y=0\)

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\left(3+1\right)\left(3x^2+y^2\right)\ge\left(3x+y\right)^2\)

\(\Rightarrow4\left(3x^2+y^2\right)\ge\left(3x+y\right)^2\)

\(\Rightarrow4\left(3x^2+y^2\right)\ge\left(3x+y\right)^2=1^2=1\)

\(\Rightarrow M=3x^2+y^2\ge\dfrac{1}{4}\)

Đẳng thức xảy ra khi \(x=y=\dfrac{1}{4}\)

Bạn nhân 4 lên rồi tách ra hằng đẳng thức

Ta có 

A=x²+xy+y²-3x-3y+2016

=>4A=4x²+4xy+y² -6(2x+y) + 9 + 3(y²-2y+1) +8052

         =(2x+y)²-6(2x+y)+9 + 3(y-1)² +8052 

        =(2x+y-3)²+3(y-1)²+8052>= 8052

     =>A>=2013

Dấu bang xay ra khi x=y=1

Ta có A= x²+xy+y²+3x-3y+2016

=> 2A= 2x²+2xy+2y²+6x-6y+4032

=> 2A=(x²+2xy+y²)+(x²+6x+9)+(y²-6y+9)+ 4014

=> 2A= (x+y)²+ (x+3)²+(y-3)²+4014

=> 2A >= 4014=> A>=2007

Dấu "=" xảy ra khi x=-3; y=-3