一道比较难的不等式求证明!a^(n/2)+b^(n/2)+c^(n/2)>=3((a+b+c)/3)^(n/2)n属于正整数是一道不等式题里的答案的过程,没法理解,另附原题:设a,b,c是三角形三边的长,且a+b+c=2S,求证:(a^n)/b+c +(b^n)/a+c +(c^
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![一道比较难的不等式求证明!a^(n/2)+b^(n/2)+c^(n/2)>=3((a+b+c)/3)^(n/2)n属于正整数是一道不等式题里的答案的过程,没法理解,另附原题:设a,b,c是三角形三边的长,且a+b+c=2S,求证:(a^n)/b+c +(b^n)/a+c +(c^](/uploads/image/z/13327202-2-2.jpg?t=%E4%B8%80%E9%81%93%E6%AF%94%E8%BE%83%E9%9A%BE%E7%9A%84%E4%B8%8D%E7%AD%89%E5%BC%8F%E6%B1%82%E8%AF%81%E6%98%8E%21a%5E%28n%2F2%29%2Bb%5E%28n%2F2%29%2Bc%5E%28n%2F2%29%3E%3D3%28%EF%BC%88a%2Bb%2Bc%EF%BC%89%2F3%29%5E%28n%2F2%29n%E5%B1%9E%E4%BA%8E%E6%AD%A3%E6%95%B4%E6%95%B0%E6%98%AF%E4%B8%80%E9%81%93%E4%B8%8D%E7%AD%89%E5%BC%8F%E9%A2%98%E9%87%8C%E7%9A%84%E7%AD%94%E6%A1%88%E7%9A%84%E8%BF%87%E7%A8%8B%2C%E6%B2%A1%E6%B3%95%E7%90%86%E8%A7%A3%2C%E5%8F%A6%E9%99%84%E5%8E%9F%E9%A2%98%EF%BC%9A%E8%AE%BEa%2Cb%2Cc%E6%98%AF%E4%B8%89%E8%A7%92%E5%BD%A2%E4%B8%89%E8%BE%B9%E7%9A%84%E9%95%BF%2C%E4%B8%94a%2Bb%2Bc%3D2S%2C%E6%B1%82%E8%AF%81%EF%BC%9A%28a%5En%29%2Fb%2Bc+%2B%28b%5En%29%2Fa%2Bc+%2B%28c%5E)
一道比较难的不等式求证明!a^(n/2)+b^(n/2)+c^(n/2)>=3((a+b+c)/3)^(n/2)n属于正整数是一道不等式题里的答案的过程,没法理解,另附原题:设a,b,c是三角形三边的长,且a+b+c=2S,求证:(a^n)/b+c +(b^n)/a+c +(c^
一道比较难的不等式求证明!
a^(n/2)+b^(n/2)+c^(n/2)>=3((a+b+c)/3)^(n/2)
n属于正整数
是一道不等式题里的答案的过程,没法理解,
另附原题:设a,b,c是三角形三边的长,且a+b+c=2S,求证:(a^n)/b+c +(b^n)/a+c +(c^n)/a+b >=((2/3)^(n-2) )* s^(n-1)
一道比较难的不等式求证明!a^(n/2)+b^(n/2)+c^(n/2)>=3((a+b+c)/3)^(n/2)n属于正整数是一道不等式题里的答案的过程,没法理解,另附原题:设a,b,c是三角形三边的长,且a+b+c=2S,求证:(a^n)/b+c +(b^n)/a+c +(c^
利用3个结论:
(1)切比雪夫不等式:
若 a1 >= a2 >= ...>= an,b1 >= b2 >= ...>= bn
则:n*(a1b1 + a2b2 + ...+ anbn) >= (a1+a2+...+an)(b1+b2+...+bn)
(2)平均不等式:
(a^n + b^n + c^n)/3 >= [ (a + b + c)/3 ]^n
(3)
a/(b+c) + b/(a+c) + c/(a+b) >= 3/2
这个先证明一下:
a/(b+c) + b/(a+c) + c/(a+b)
= (a+b+c)/(a+b) + (a+b+c)/(b+c) + (a+b+c)/(c+a) - 3
= (a + b + c) * ( 1/(a+b) + 1/(b+c) + 1/(c+a) ) - 3
= 0.5 * (a+b+b+c+c+a)*[ 1/(a+b) + 1/(b+c) + 1/(c+a) ] - 3
>= 0.5 * { 3*[(a+b)(b+c)(c+a)]^1/3 } * { 3*[1/(a+b)*1/(b+c)*1/(c+a)]^1/3 } - 3
= 0.5 * 3 * 3 - 3
= 3/2
原题目的证明如下:
根据a、b、c的对称性,不妨假设 a >= b >= c,则:
1/(b+c) >= 1/(a+c) >= 1/(a+b)
因此:
a^(n-1) >= b^(n-1) >= c^(n-1)
a/(b+c) >= b/(a+c) >= c/(a+b)
根据切比雪夫不等式,有:
a^n/(b+c)+b^n/(a+c)+c^n/(a+b)
= a^(n-1)*(a/(b+c)) + b^(n-1)*(b/(a+c)) + c^(n-1)*(c/(a+b))
>= 1/3 * ( a^(n-1) + b^(n-1) + c^(n-1) ) ( a/(b+c) + b/(a+c) + c/(a+b) )
>= [ (a + b + c)/3 ]^(n-1) * ( a/(b+c) + b/(a+c) + c/(a+b) )
= (2s/3)^(n-1) * ( a/(b+c) + b/(a+c) + c/(a+b) )
>= (2s/3)^(n-1) * 3/2
= (2/3)^(n-2)*s^(n-1)
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