已知x,y,z,均大于0,求证x²/a+y²/b+z²/c≥ (x+y+z)²/(a+b+c)
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![已知x,y,z,均大于0,求证x²/a+y²/b+z²/c≥ (x+y+z)²/(a+b+c)](/uploads/image/z/13352939-35-9.jpg?t=%E5%B7%B2%E7%9F%A5x%2Cy%2Cz%2C%E5%9D%87%E5%A4%A7%E4%BA%8E0%2C%E6%B1%82%E8%AF%81x%26%23178%3B%2Fa%2By%26%23178%3B%2Fb%2Bz%26%23178%3B%2Fc%E2%89%A5+%28x%2By%2Bz%29%26%23178%3B%2F%28a%2Bb%2Bc%29)
已知x,y,z,均大于0,求证x²/a+y²/b+z²/c≥ (x+y+z)²/(a+b+c)
已知x,y,z,均大于0,求证x²/a+y²/b+z²/c≥ (x+y+z)²/(a+b+c)
已知x,y,z,均大于0,求证x²/a+y²/b+z²/c≥ (x+y+z)²/(a+b+c)
(a+b+c)(x²/a+y²/b+z²/c)≥(x+y+z)²
所以(x²/a+y²/b+z²/c)≥(x+y+z)²/(a+b+c)
柯西不等式