Question

已知 x ;y; z＞0 x+y+z=3 求 1/x+1/y+1/z最小值 2.证明3≤x2+y2+z2＜9

Answer 1

(1)
依Cauchy不等式得
1/x+1/y+1/z
=1²/x+1²/y+1²/z
≥(1+1+1)²/(x+y+z)
=9/3
=3.
故x=y=z=1时，
所求最小值为3.

(2)
x²+y²+z²
=x²/1+y²/1+z²/1
≥(x+y+z)²/(1+1+1)
=9/3
=3，
x²+y²+z²
=(x+y+z)²-2(xy+yz+zx)
<(x+y+z)²
=9，
∴3≤x²+y²+z²<9。