Question

Cho x + y = 2. Chứng minh rằng x.y \(\le\)1.

 

Answer 1

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 \(\Rightarrow\)\(2\ge2\sqrt{xy}\)\(\Rightarrow1\sqrt{xy}\le1\)\(\Rightarrow xy\le1\)

Answer 2

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

Answer 3

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