已知x,y∈R且x+y=1,求x^2+y^2的最小值

(x-y)^2>=0
X^2+Y^2>=2xy
2(X^2+Y^2)>=X^2+Y^2+2xy=(x+y)^2=1
所以X^2+Y^2>=1/2,最小值是1/2
=================================================================================x+y=1
(x+y)^2=x^2+y^2+2*x*y=1
x^2+y^2=1-2xy;
显然x=y时有最小值，
又x+y=1.
即x=y=1/2;
x^2+y^2=1/2;

x^2+y^2=x^2+(1-x)^2=2x^2-2x+1=2(x^2-x+1/2)=2(x-1/2)^2+1/2 x属于实数，
所以x^2+y^2的最小值为1/2

x+y=1 y=1-x
x^2+y^2=（x+y)^2-2xy=1-2x(1-x)=2x^2-2x+1=2(x-1/2)^2-(1/2)^2+1=2(x-1/2)^2+3/4
开口向上，当x=1/2时，x^2+y^2有最小值为3/4

2

x=1-y代入得
(1-y)²+y²
=2y²-2y+1
=2(y-1/2)²+1/2
当y=1/2时有最小值1/2