$a,b,c>0,a+b+c=21$ prove that $a+\sqrt{ab} +\sqrt[3]{abc} \leq 28$
I have tried to use AM-GM inequality, but get no result as follows: $$a+\sqrt{ab}+\sqrt[3]{abc}\leq a+\frac{a+b}{2}+\frac{a+b+c}{3}$$
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Source of the problem, please? – Gerry Myerson Aug 27 '13 at 12:56
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1Equality, by the way, is obtained when $a=16$, $b=4$, $c=1$. – Gerry Myerson Aug 27 '13 at 12:58
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How to prove it or how to get it. it's a problem in a book for Mathematics Contest. – user89007 Aug 27 '13 at 13:19
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1What book, please? What page? – Gerry Myerson Aug 27 '13 at 13:20
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1蔡玉书《数学奥林匹克不等式证明方法和技巧》,a exercise. – user89007 Aug 27 '13 at 13:22
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Hmm. Is there anything in there that might give a clue as to the method to use? Does the book give a source for the problem? – Gerry Myerson Aug 27 '13 at 13:25
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could you give me a clue. I could try it myself. nothing in the book. – user89007 Aug 27 '13 at 13:27
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related to 2^n? I find $16=2^4,4=2^2,1=2^0$ – user89007 Aug 27 '13 at 13:31
1 Answers
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$a+\sqrt{ab}+\sqrt[3]{abc}=a+\frac{\sqrt{4ab}}{2}+\frac{\sqrt[3]{64abc}}{4}\le a+\frac{a+4b}{4}+\frac{a+4b+16c}{12}=\frac{4(a+b+c)}{3}=28$
Steven Sun
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Thx.I have understand more about the application of the AM-GM equality. – user89007 Aug 27 '13 at 14:05