来讨论一道简单不等式题.已知x+2y=1 求x^2+y^2最小值.我的问题在于用柯西不等式算为:(x^2+y^2)(1^2+2^2)大于等于x+2y.解得x^2+y^2最小值为1/5。可是按照均值不等式算当x^2=y^2时x^2+y^2有最小值.即x=y,带
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/07 14:22:49
![来讨论一道简单不等式题.已知x+2y=1 求x^2+y^2最小值.我的问题在于用柯西不等式算为:(x^2+y^2)(1^2+2^2)大于等于x+2y.解得x^2+y^2最小值为1/5。可是按照均值不等式算当x^2=y^2时x^2+y^2有最小值.即x=y,带](/uploads/image/z/2770081-25-1.jpg?t=%E6%9D%A5%E8%AE%A8%E8%AE%BA%E4%B8%80%E9%81%93%E7%AE%80%E5%8D%95%E4%B8%8D%E7%AD%89%E5%BC%8F%E9%A2%98.%E5%B7%B2%E7%9F%A5x%2B2y%3D1+%E6%B1%82x%5E2%2By%5E2%E6%9C%80%E5%B0%8F%E5%80%BC.%E6%88%91%E7%9A%84%E9%97%AE%E9%A2%98%E5%9C%A8%E4%BA%8E%E7%94%A8%E6%9F%AF%E8%A5%BF%E4%B8%8D%E7%AD%89%E5%BC%8F%E7%AE%97%E4%B8%BA%3A%28x%5E2%2By%5E2%29%281%5E2%2B2%5E2%29%E5%A4%A7%E4%BA%8E%E7%AD%89%E4%BA%8Ex%2B2y.%E8%A7%A3%E5%BE%97x%5E2%2By%5E2%E6%9C%80%E5%B0%8F%E5%80%BC%E4%B8%BA1%2F5%E3%80%82%E5%8F%AF%E6%98%AF%E6%8C%89%E7%85%A7%E5%9D%87%E5%80%BC%E4%B8%8D%E7%AD%89%E5%BC%8F%E7%AE%97%E5%BD%93x%5E2%3Dy%5E2%E6%97%B6x%5E2%2By%5E2%E6%9C%89%E6%9C%80%E5%B0%8F%E5%80%BC.%E5%8D%B3x%3Dy%2C%E5%B8%A6)
xSMP+,[[eiLąKCtSv/Çf
G?^ە*b
;{{=TN%>g뮡JwZۚVYdsKKy
k[?W~ˀlumIw
]!h!s<Ϣ"|H(`Z7-:&(j\KnnBs`H"[}TtC[ӳ:' (ApDdN#p%ϳ).{OBBA^0(%%ZcۍT٢<V8s@
X"TxYH%uTϔ0%H