设x,y,z为不全为零的实数,求证 (2yz+2zx+xy)/(x^2+y^2+z^2)≤(√33+1)/4设x,y,z为不全为零的实数,求证(2yz+2zx+xy)/(x^2+y^2+z^2)≤(√33+1)/4 证明:只需考虑x≥0,y≥0,z≥0,2yz+2zx+xy≤1/2xx+1/2yy+γyy+(1/γ)zz+(1/γ)zz+γx
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![设x,y,z为不全为零的实数,求证 (2yz+2zx+xy)/(x^2+y^2+z^2)≤(√33+1)/4设x,y,z为不全为零的实数,求证(2yz+2zx+xy)/(x^2+y^2+z^2)≤(√33+1)/4 证明:只需考虑x≥0,y≥0,z≥0,2yz+2zx+xy≤1/2xx+1/2yy+γyy+(1/γ)zz+(1/γ)zz+γx](/uploads/image/z/10231797-21-7.jpg?t=%E8%AE%BEx%2Cy%2Cz%E4%B8%BA%E4%B8%8D%E5%85%A8%E4%B8%BA%E9%9B%B6%E7%9A%84%E5%AE%9E%E6%95%B0%2C%E6%B1%82%E8%AF%81+%282yz%2B2zx%2Bxy%29%2F%28x%5E2%2By%5E2%2Bz%5E2%29%E2%89%A4%28%E2%88%9A33%2B1%29%2F4%E8%AE%BEx%2Cy%2Cz%E4%B8%BA%E4%B8%8D%E5%85%A8%E4%B8%BA%E9%9B%B6%E7%9A%84%E5%AE%9E%E6%95%B0%2C%E6%B1%82%E8%AF%81%282yz%2B2zx%2Bxy%29%2F%28x%5E2%2By%5E2%2Bz%5E2%29%E2%89%A4%28%E2%88%9A33%2B1%29%2F4+%E8%AF%81%E6%98%8E%EF%BC%9A%E5%8F%AA%E9%9C%80%E8%80%83%E8%99%91x%E2%89%A50%2Cy%E2%89%A50%2Cz%E2%89%A50%2C2yz%2B2zx%2Bxy%E2%89%A41%2F2xx%2B1%2F2yy%2B%CE%B3yy%2B%281%2F%CE%B3%29zz%2B%281%2F%CE%B3%29zz%2B%CE%B3x)
设x,y,z为不全为零的实数,求证 (2yz+2zx+xy)/(x^2+y^2+z^2)≤(√33+1)/4设x,y,z为不全为零的实数,求证(2yz+2zx+xy)/(x^2+y^2+z^2)≤(√33+1)/4 证明:只需考虑x≥0,y≥0,z≥0,2yz+2zx+xy≤1/2xx+1/2yy+γyy+(1/γ)zz+(1/γ)zz+γx
设x,y,z为不全为零的实数,求证 (2yz+2zx+xy)/(x^2+y^2+z^2)≤(√33+1)/4
设x,y,z为不全为零的实数,求证
(2yz+2zx+xy)/(x^2+y^2+z^2)≤(√33+1)/4
证明:只需考虑x≥0,y≥0,z≥0,
2yz+2zx+xy≤1/2xx+1/2yy+γyy+(1/γ)zz+(1/γ)zz+γxx……
为什么要这样设γ?是怎么想到的?
设x,y,z为不全为零的实数,求证 (2yz+2zx+xy)/(x^2+y^2+z^2)≤(√33+1)/4设x,y,z为不全为零的实数,求证(2yz+2zx+xy)/(x^2+y^2+z^2)≤(√33+1)/4 证明:只需考虑x≥0,y≥0,z≥0,2yz+2zx+xy≤1/2xx+1/2yy+γyy+(1/γ)zz+(1/γ)zz+γx
这个主要是由於分子和分母是齐次的,而且yz和xz系数相同,但是与xy不相同,所以要添加一个系数让
1/2(x²+y²) >=xy
γx² + 1/γ z² >= 2xz
γy² + 1/γ z² >= 2yz
相加得(1/2+1/γ)x² + (1/2+1/γ)y² + 2/γz² >= 2yz+2zx+xy
若(1/2+1/γ) = 2/γ = 1刚好得γ = 2
其实就是刚好凑出来得,当然可以分别给3个方程都配上系数,那麽可以得到一个3元1次方程,可以解出3个系数来,和这裏得系数刚好一样