-(y^2-2y+1)-3=-(y-1)^2-3=<-3<0

Max=-3 <=>y=1

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\(-y^2+2y-4=-\left(y^2-2y+1\right)-3=-\left(y-1\right)^2-3< 0\)

=> đpcm

5x^2+2y^2+4xy-4x-y+5=(4x^2+y^2+4xy)+(x^2-4x+4)+(y^2-y+1/4)+3/4 =(2x+y)^2+(x-2)^2+(y-1/2)^2+3/4  (1)

 vi (2x+y)^2>=0 , (x-2)^2>=0  ,(y-1/2)^2>=0 (2)

tu 1 va 2 suy ra dieu phai chung minh

(x+y)^2+(x+2)^2-(-x-y)^2+x^2+y^2+1>=0

 x² + y² + xy - 2x - 2y + 2 

= (x² - 2x + 1) + (xy - y) + y²/4 + 3y²/4 - y + 1/3 + 2/3 

= [ (x - 1)² + 2.(x - 1).y/2 + y²/4 ] + 3.[ (y/2)² - 2.y/2.1/3 + 1/9 ] + 2/3 

= (x - 1 + y/2)² + 3(y/2 - 1/3)² + 2/3 

có: 
(x - 1 + y/2)² ≥ 0 
3(y/2 - 1/3)² ≥ 0 

--> (x - 1 + y/2)² + 3(y/2 - 1/3)² + 2/3 > 0 

hay x² + y² + xy - 2x - 2y + 2 > 0 --> đ.p.c.m

\(x^2+y^2-x+2y+4\)

\(=x^2+y^2-x+2y+1+\frac{12}{4}\)

\(=x^2-x+\frac{1}{4}+y^2+2y+1+\frac{11}{4}\)

\(=\left(x-\frac{1}{2}\right)^2+\left(y+1\right)^2+\frac{11}{4}\)

Dễ thấy :\(\hept{\begin{cases}\left(x-\frac{1}{2}\right)^2\ge0\\\left(y+1\right)^2\ge0\end{cases}}\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y+1\right)^2\ge0\)

\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y+1\right)^2+\frac{11}{4}\ge\frac{11}{4}>0\)

Ta có : x² + 2x + 2

= x² + 2x + 1 + 1

= (x + 1)² + 1 \(\ge1\forall x\)

Vậy  x² + 2x + 2 \(>0\forall x\)

Ta có : x² + 2x + 2

=> x² + 2x + 1 + 1

=> ( x + 1)² + 1  >  1\(\forall x\)

Vậy x² + 2x + 2   > \(0\forall x\)

đùa nhau

Ta có : \(x^2+y^2-2x-2y+2017\)

\(=\left(x^2-2x+1\right)+\left(y^2-2y+1\right)+2015\)

\(=\left(x-1\right)^2+\left(y-1\right)^2+2015\)

Vì : \(\left(x-1\right)^2\ge0\forall x\in R\) ; \(\left(y-1\right)^2\ge0\forall x\in R\)

Nên : \(\left(x-1\right)^2+\left(y-1\right)^2+2015\ge0+0+2015=2015>0\forall x\in R\)

Vậy \(x^2+y^2-2x-2y+2017\ge0\forall x\in R\)

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

\(=\left(x-\frac{2y+1}{2}\right)^2+\frac{1}{4}\left(2y-1\right)^2+\frac{1}{2}>0\)

bn có thể lm rõ hơn dc chứ

\(\Leftrightarrow x^2-2.3.x+9+1=\left(x-3\right)^2+1\Rightarrow\hept{\begin{cases}\left(x-3\right)^2\ge0\\1>0\end{cases}}\Rightarrow\left(x-3\right)^2+1>0\)

\(\Leftrightarrow x^2-2.\frac{3}{2}.x+\frac{9}{4}+\frac{7}{4}=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\Leftrightarrow\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0\\\frac{7}{4}>0\end{cases}}\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{7}{4}>0\)

\(\Leftrightarrow2.\left(x^2+xy+y^2+1\right)=x^2+2xy+y^2+x^2+y^2+2=\left(x+y\right)^2+x^2+y^2+2\)

ta có \(\left(x+y\right)^2\ge0,x^2\ge0,y^2\ge0,2>0\Rightarrow\left(x+y\right)^2+x^2+y^2+2>0\)

\(\Leftrightarrow x^2-2xy+y^2+x^2-2.1x+1+y^2+2.2.y+4+3\)\(=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3\)

Ta có \(=\left(x-y\right)^2\ge0,\left(x-1\right)^2\ge0,\left(y+2\right)^2\ge0,3>0\)\(\Rightarrow=\left(x-y\right)^2+\left(x-1\right)^2+\left(y+2\right)^2+3>0\)

T i c k cho mình 1 cái nha mới bị trừ 50 đ

Do \(x,y>0\) BĐT tương đương:

\(\dfrac{x^2+2y^2+3}{2}\ge xy+y+1\)

\(\Leftrightarrow x^2+2y^2+3\ge2xy+2y+2\)

\(\Leftrightarrow x^2-2xy+y^2+y^2-2y+1\ge0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-1\right)^2\ge0\) (luôn đúng)

Vậy BĐT được chứng minh xong

Vì x,y>0 nên các mẫu thức dương.

BĐT<=>\(2\left(xy+y+1\right)\le x^2+2y^2+3\\ \Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)\ge0\\ \Leftrightarrow\left(x-y\right)^2+\left(y-1\right)^2\ge0\left(1\right)\)

(1) đúng với mọi x,y>0 nên BĐT đã cho được chứng minh.

Dấu "=" xảy ra khi và chỉ khi x=y=1.