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If we have $20$ real positive number $a_1, a_2, \ldots, a_{20}$ then how do I prove that $\frac{a_1}{a_4} + \frac{a_2}{a_5} + \ldots + \frac{a_{18}}{a_1}+ \frac{a_{19}}{a_2}+\frac{a_{20}}{a_3} \gt 20$.

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  • $\begingroup$ Just apply AM-GM, noting that the product of all the fractions is $1$. $\endgroup$ – Quang Hoang Oct 6 '15 at 6:20
  • $\begingroup$ what exactly do you mean? the product of all the fraction is 1, but how do I use am-gm to prove it? isn't am-gm mean that the additive porduct is greater than the geometic product. How does that apply $\endgroup$ – lodnots3 Oct 6 '15 at 6:25
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Hint: The product of the terms is $1$. Use the Arithmetic Mean/ Geometric Mean Inequality.

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  • $\begingroup$ what do you mean the product of the term is 1. as in a1/a4 * a2/5 etc is 1? $\endgroup$ – lodnots3 Oct 6 '15 at 6:24
  • $\begingroup$ Every $a_i$ occurs once in a numerator and once in a denominator. $\endgroup$ – André Nicolas Oct 6 '15 at 6:28
  • $\begingroup$ okay, so how do I use arithmetic mean/gometric mean inequality? I know that arithmetic mean/geometric mean inequality mean that the the arithmetic mean is greater than the geometric mean. $\endgroup$ – lodnots3 Oct 6 '15 at 6:31
  • $\begingroup$ The product of the terms is $\frac{a_1a_2a_3a_4\cdots a_{20}}{a^4a_5\cdots a_{20}a_1a_2a_3}$. Everything cancels, we get $1$. $\endgroup$ – André Nicolas Oct 6 '15 at 6:32
  • $\begingroup$ To use AM/GM, let our fractions be $x_1$ to $x_{20}$. Then by AM/GM we have $\frac{x_1+\cdots+x_{20}}{20}\ge (x_1x_2\cdots x_{20})^{1/20}=1$. $\endgroup$ – André Nicolas Oct 6 '15 at 6:34

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