Question

设x1,x2及y1,y2使为实数使得(x1)2+(x2)2≤1证明不等式

Answer 1

y+z=a-x,y^2+z^2=a^2-x^2,

2yz=(a-x)^2-(a^2-x^2)=2x^2-2ax,yz=(x^2-ax)

所以y,z是方程t^2-(a-x)t+(x^2-ax)=0的两个根.

判别式=(a-x)^2-4(x^2-ax)=-3x^2+2ax+a^2>=0

解得-3/a