用轮换对称式解决求最值得问题x>0,y>0.则(x/2x+y)+(y/x+2y)的最大值为?
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![用轮换对称式解决求最值得问题x>0,y>0.则(x/2x+y)+(y/x+2y)的最大值为?](/uploads/image/z/9498984-24-4.jpg?t=%E7%94%A8%E8%BD%AE%E6%8D%A2%E5%AF%B9%E7%A7%B0%E5%BC%8F%E8%A7%A3%E5%86%B3%E6%B1%82%E6%9C%80%E5%80%BC%E5%BE%97%E9%97%AE%E9%A2%98x%3E0%2Cy%3E0.%E5%88%99%28x%2F2x%2By%29%2B%28y%2Fx%2B2y%29%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BA%3F)
用轮换对称式解决求最值得问题x>0,y>0.则(x/2x+y)+(y/x+2y)的最大值为?
用轮换对称式解决求最值得问题
x>0,y>0.则(x/2x+y)+(y/x+2y)的最大值为?
用轮换对称式解决求最值得问题x>0,y>0.则(x/2x+y)+(y/x+2y)的最大值为?
x/(2x+y)+y/(x+2y) = 1-(x+y)/(2x+y)+1-(x+y)/(x+2y)
= 2-(x+y)(1/(2x+y)+1/(x+2y))
= 2-1/3·((2x+y)+(x+2y))(1/(2x+y)+1/(x+2y))
≤ 2-1/3·(1+1)² (Cauchy不等式)
= 2/3.
可知x = y时等号成立,即最大值为2/3.