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

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

Bài 1 : 

a) \(x^2+y^2\)

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

\(\Leftrightarrow\left(x+y\right)^2-2xy=\left(-3\right)^2-2.\left(-28\right)=65\)

b) \(x^3+y^3\)

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

\(\Leftrightarrow\left(x+y\right)\left(x^2+2xy+y^2-3xy\right)\)

\(\Leftrightarrow\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]=\left(-3\right)\left[\left(-3\right)^2-3.\left(-28\right)\right]=-279\)

c) \(x^4+y^4\)

\(\Leftrightarrow\left(x+y\right)^4-4x^3y-4xy^3-6x^2y^2=\left(-3\right)^4-4\left(-28\right).65-6\left(-28\right)^2=2657\)

Bài 3:

Có:    \(x^3+y^3+z^3=\left(x+y\right)^3-3xy\left(x+y\right)+z^3\)

=>     \(x^3+y^3+z^3=\left(-z\right)^3-3xy.-z+z^3\)

=>     \(x^3+y^3+z^3=-z^3+z^3+3xyz=3xyz\)

=> TA CÓ ĐPCM.

VẬY      \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)

Bài 2 :

a) Giả sử  \(a^2+b^2+c^2+d^2\ge ab+ac+ad\)

\(\Leftrightarrow a^2+b^2+c^2+d^2-ab-ac-ad\ge0\)

\(\Leftrightarrow4a^2+4b^2+4c^2+4d^2-4ab-4ac-4ad\ge0\)

\(\Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d^2\right)\ge0\)( luôn đúng )

\(\RightarrowĐPCM\)

b) Sửa đề : \(a^2+4b^2+4c^2\ge2ab-2ac+4bc\)

Ta có : \(\left(a+2c\right)^2\ge0\Leftrightarrow a^2+4c^2\ge-4ac\left(1\right)\)

Áp dụng BĐT Cô - si ta có :

\(\hept{\begin{cases}a^2+4b^2\ge4ab\left(2\right)\\4b^2+4c^2\ge8bc\left(3\right)\end{cases}}\)

(1) + (2) + (3) 

\(\Leftrightarrow2a^2+8b^2+8c^2\ge4ab-4ac+8bc\)

\(\Leftrightarrow2\left(a^2+4b^2+4c^2\right)\ge4\left(ab-ac+2bc\right)\)

\(\Leftrightarrow a^2+4b^2+4c^2\ge2ab-2ac+4bc\)