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: \(4xy\le\left(x+y\right)^2\)

Lại có: \(x;y>0\)

\(\Rightarrow\left(x+y\right)^2xy>0\)

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

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

Ta có :

\(\left(x+y\right)^2-4xy\)

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

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

\(=\left(x-y\right)^2\ge0\)

\(\Rightarrow\left(x+y\right)^2\ge4xy\)

Lại có : \(x,y>0\)

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

\(\Rightarrow\frac{4}{\left(x+y\right)^2}\le\frac{1}{xy}\)<đpcm>

  x+y=2 
<=> x=2-y(1) 
giả sử x*y≤1 
<=>(2-y)y≤1 
<=>y^2 - 2y +1≥0 
<=> (y-1)^2≥0 
<=>y≥1(2) 
từ (1),(2)=> x*y≤1 
 

1/ Ta có :

\(x+y=2\)

\(\Leftrightarrow x=2-y\)

\(\Leftrightarrow xy=y\left(2-y\right)\)

\(\Leftrightarrow xy=2y-y^2\)

\(\Leftrightarrow xy=-y^2+2y-1+1\)

\(\Leftrightarrow xy=-\left(y-1\right)^2+1\)

Với mọi x ta có :

\(\left(y-1\right)^2\ge0\)

\(-\left(y-1\right)^2\le0\)

\(\Leftrightarrow-\left(y-1\right)^2+1\le1\)

\(\Leftrightarrow xy\le1\left(đpcm\right)\)

2/ Ta có :

\(E=\dfrac{x^2+8}{x^2+2}=\dfrac{x^2+2+6}{x^2+2}=\dfrac{x^2+2}{x^2+2}+\dfrac{6}{x^2+2}=1+\dfrac{6}{x^2+2}\)

Để E lớn nhất thì \(\dfrac{6}{x^2+2}\) đạt GTLN

\(\Leftrightarrow x^2+2\) đạt GTNN

\(\Leftrightarrow x^2+2=1\)

\(\Leftrightarrow x^2=-1\)

\(\Leftrightarrow x\in\varnothing\)

Vậy ....

1)Ta có:\(\left(x-y\right)^2\ge0\forall x,y\in R\)

\(\Rightarrow x^2-2xy+y^2\ge0\)

\(\Rightarrow x^2+2xy+y^2-4xy\ge0\)

\(\Rightarrow\left(x+y\right)^2\ge4xy\)

\(\Rightarrow4xy\le2^2=4\)

\(\Rightarrow xy\le1\left(đpcm\right)\)

2)Ta có:\(x^2\ge0\)

\(\Rightarrow x^2+2\ge2\)

\(\Rightarrow\dfrac{6}{x^2+2}\le\dfrac{6}{2}=3\)

Áp dụng: \(E=\dfrac{x^2+8}{x^2+2}\)

\(E=\dfrac{x^2+2+6}{x^2+2}\)

\(E=1+\dfrac{6}{x^2+2}\)

\(E\le1+3=4\)

\(\Rightarrow MAXE=4\Leftrightarrow x=0\)

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