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| − | == Problem ==
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| − | Let <math>{{a}_{2}}, {{a}_{3}}, \cdots, {{a}_{n}}</math> be positive real numbers that satisfy <math>{{a}_{2}}\cdot {{a}_{3}}\cdots {{a}_{n}}=1</math> . Prove that
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| − | <cmath> \left(a_2+1\right)^2\cdot \left(a_3+1\right)^3\cdots \left(a_n+1\right)^n\gneq n^n</cmath>
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| − | == Solution ==
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| − | The inequality between arithmetic and geometric mean implies
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| − | <cmath>{{\left( {{a}_{k}}+1 \right)}^{k}}={{\left( {{a}_{k}}+\frac{1}{k-1}+\frac{1}{k-1}+\cdots +\frac{1}{k-1} \right)}^{k}}\ge {{k}^{k}}\cdot {{a}_{k}}\cdot \frac{1}{{{\left( k-1 \right)}^{k-1}}}=\frac{{{k}^{k}}}{{{\left( k-1 \right)}^{k-1}}}\cdot {{a}_{k}}</cmath>
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| − | The inequality is strict unless <math>a_k=\frac1{k-1}</math>. Multiplying analogous inequalities for <math>k=2,\text{ 3, }\cdots \text{, }n</math> yields
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| − | <cmath>\left(a_2+1\right)^2\cdot \left(a_3+1\right)^3\cdots \left(a_n+n\right)^n\gneq \frac{2^2}{1^1}\cdot\frac{3^3}{2^2}\cdot \frac{4^4}{3^3}\cdots \frac{n^n}{(n-1)^{n-1}}\cdot a_2a_3\cdots a_n=n^n</cmath>
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