If $a_i>0$ and $\sum_{i=1}^n a_i = 1$, is $\sum_{i=1}^n \frac{1}{a_i} \geq n^2$? I'm doing an inequality exercise. If I can confirm that's true, then my proof is done. I wrote down some examples and they are all true. I guess we need to compare each $\frac{1}{a_i}$ with $\frac{1}{n}$.
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$\begingroup$ Are the $a_i>0?$ $\endgroup$– Adrian KeisterMay 17, 2018 at 14:47
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1$\begingroup$ @AdrianKeister Right. I edited it. $\endgroup$– user547265May 17, 2018 at 14:49
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2$\begingroup$ Hint: $\sqrt a(1/\sqrt a)=1$. Cauchy--Schwarz. $\endgroup$– David C. UllrichMay 17, 2018 at 14:50
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3 Answers
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HM-AM says
$$\frac{n}{\displaystyle\sum_{k=1}^n\frac{1}{a_k}} \leq \frac{1}{n}\sum_{k=1}^na_k.$$
So, you are done.
answered May 17, 2018 at 14:50
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$$\sum_{i=1}^n\frac{1}{a_i}-n^2=\sum_{i=1}^n\left(\frac{1}{a_i}-n\right)=\sum_{i=1}^n\left(\frac{1-na_i}{a_i}+n(na_i-1)\right)=\sum_{i=1}^n\frac{(na_i-1)^2}{a_i}\geq0.$$
answered May 17, 2018 at 15:08
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1$\begingroup$ (+1) Glad to see an answer that doesn't rely on referring to a known general inequality, but rather simply uses straightforward arithmetic. Well done. $\endgroup$ May 17, 2018 at 15:39
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Use that $$\frac{a_1+a_2+...+a_n}{n}\geq \frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}}$$ for all $$a_i>0,i=1...n$$
answered May 17, 2018 at 14:52