x+y=2

=> (x+y)²=4

=> x²+y²+2xy = 4

Áp dụng x²+y² >= 2xy   

=> x²+y²+2xy >= 4xy

Mà x²+y²+2xy = 4

=> 4>= 4xy

=> xy <= 1

x+y=2

\(\Rightarrow\)x=1; x=0; x=-1; x=-2;...

y=1; y=2; y=3; y=4;...

\(\Rightarrow\)x.y= 1.1=1=1

0.2=0<1

-1.3=-3<1

-2.4=-8<1

.............

\(\Rightarrow\)Nếu x+y=2 thì x.y\(\le\)1

Ta có: \(x+y=2\)

\(\Rightarrow x=2-y.\)

Có: \(x.y=\left(2-y\right).y\)

\(\Rightarrow x.y=2y-y^2\)

\(\Rightarrow x.y=-y^2+2y-1+1\)

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

Vì \(\left(y-1\right)^2\ge0\) \(\forall y.\)

\(\Rightarrow-\left(y-1\right)^2\le0\) \(\forall y.\)

\(\Rightarrow-\left(y-1\right)^2+1\le1\) \(\forall y.\)

\(\Rightarrow x.y\le1\left(đpcm\right).\)

Chúc bạn học tốt!

Đề của bạn thiếu dấu bằng.

Ta có: 

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

Dấu "=" xảy ra <=> x = y = 1/2

sai rồi

giả sử\(\left|x.y\right|=\left|x\right|.\left|y\right|\)

 \(\Leftrightarrow\left|x.y\right|:\left|x\right|=y\)

\(\Rightarrow\left|x\right|=\left|y\right|\)

\(\Rightarrow\left|xy\right|=\left|x\right|.\left|y\right|\)

suy ra điều phải chứng minh 

\(\left|x.y\right|=\left|x\right|.\left|y\right|\)

\(\Leftrightarrow\left|x.y\right|:\left|x\right|=\left|y\right|\)

\(\Leftrightarrow\left|y\right|=\left|y\right|\)

Vậy \(\left|x.y\right|=\left|x\right|.\left|y\right|\)

Ta có \(\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\)\(\Rightarrow x-2\sqrt{xy}+y\ge0\)\(\Rightarrow x+y\ge2\sqrt{xy}\)

Mà x + y = 2 \(\Rightarrow\)\(2\ge2\sqrt{xy}\)\(\Rightarrow1\sqrt{xy}\le1\)\(\Rightarrow xy\le1\)

 Vi 2 = 2 + 0 ; 1 + 1 .nen x.y = 2 . 0 ; 1.1 chi bang 0 hoac 1 nen x.y <= 1

Có : (x-y)^2 >= 0

<=> x^2+y^2-2xy >= 0

<=> 2xy < = x^2+y^2

<=> 4xy < = x^2+y^2+2xy = (x+y)^2 = 2^2 = 4

<=> xy < = 4 : 4 = 1

Dấu "=" xảy ra <=> x=y=1

=> ĐPCM

Tk mk nha

Ta có:

A = (x + 1)(y + 1)

=> A = xy + x + y +1

=> A = 1 + x + y + 1

=> A = 2 + x + y

Vì x > 0 ; y > 0

=>x \(\ge\)1; y\(\ge\)1

=> x + y \(\ge\)2

=> 2 + x + y \(\ge\)4

hay A \(\ge\)4