(x^2+y^2+z^2)*(1^2+2^2+3^2)>=(x+2y+3z)^2=1
=>x^2+y^2+z^2>=1/14
(柯西不等式)已知：x+2y+3z=1,則x^2+y^2+z^2的最小值是
(柯西不等式)已知：x+2y+3z=1,則x^2+y^2+z^2的最小值是
數(shù)學(xué)人氣：172 ℃時(shí)間：2020-01-28 14:58:45
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