- 一道不等式已知正数x,y,z满足x+y+z≥1,求证∑x^(3/
- 已知正数x, y, z满足x+y+z≥1, 求证∑x^(3/2)/(y+z)≥√3/2
- 证明：
∵x/(y+z)≥3/2·x^(3/2)/[x^(3/2)+y^(3/2)+z^(3/2)]
⇔ 3x^(1/2)·(y+z)≤2[x^(3/2)+y^(3/2)+z^(3/2)]
而[x^(3/2)+y^(3/2)+y^(3/2)]≥3x^(1/2)·y
[x^(3/2)+z^(3/2)+z^(3/2)]≥3x^(1/2)·z
∴x^(3/2)/(y+z)≥3/2·x²/[x^(3/2)+y^(3/2)+z^(3/2)]
⇔ 3/2·x²/[x^(3/2)+y^(3/2)+z^(3/2)]≥1/√3
⇔ 3(x²+y²+z²)²≥[x^(3/2)+y^(3/2)+z^(3/2)]²
⇔ 3(x²+y²+z²)²(x+y+z)≥[x^(3/2)+y^(3/2)+z^(3/2)]²(x+y+z)
∵(x²+y²+z²)(x+y+z)≥[x^(3/2)+y^(3/2)+z^(3/2)]²
3(x²+y²+z²)≥(x+y+z)²≥(x+y+z)
上两式相乘即得证.