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Prove that for $a_k>0,k=1,2,\dots,n$,

$$\sum_{k=1}^n \frac{2k+1}{a_1+a_2+\ldots+a_k}<4\sum_{k=1}^n\frac1{a_k}\;.$$

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1 Answer 1

up vote 2 down vote accepted

I must confess this problem took me a very long time!

Step1. If $a_1,a_2,\alpha,\beta,\gamma$ are positive real numbers and $\gamma=\alpha+\beta$ holds, $$\frac{\gamma^2}{a_1+a_2}\leq \frac{\alpha^2}{a_1}+\frac{\beta^2}{a_2}$$ holds too, since it is equivalent to $(\alpha a_2-\beta a_1)^2\geq 0$.

Step2. If $a_1,a_2,\alpha,\beta,\gamma,\delta$ are positive real numbers and $\delta=\alpha+\beta+\gamma$ holds, $$\frac{\delta^2}{a_1+(a_2+a_3)}\leq \frac{\alpha^2}{a_1}+\frac{(\beta+\gamma)^2}{a_2+a_3}\leq\frac{\alpha^2}{a_1}+\frac{\beta^2}{a_2}+\frac{\gamma^2}{a_3}$$ holds too, in virtue of Step2. By induction, it is easy to prove the analogous statement for $k$ variables $a_1,\ldots,a_k$. In fact, this is useless to the proof, but quite interesting in itself :)

Step3. By Step1, $$\sum_{k=1}^{n}\frac{2k+1}{a_1+\ldots+a_k}-\frac{4}{a_n}\leq \sum_{k=1}^{n-1}\frac{2k+1}{a_1+\ldots+a_k}+\frac{(\sqrt{2n+1}-2)^2}{a_1+\ldots+a_{n-1}}\leq \sum_{k=1}^{n-2}\frac{2k+1}{a_1+\ldots+a_k}+\frac{n^2}{a_1+\ldots+a_{n-1}}$$

Step4. By Step3, $$\sum_{k=1}^{n}\frac{2k+1}{a_1+\ldots+a_k}-\left(\frac{4}{a_n}+\frac{4}{a_{n-1}}\right)\leq \sum_{k=1}^{n-2}\frac{2k+1}{a_1+\ldots+a_k}+\frac{(n-2)^2}{a_1+\ldots+a_{n-2}}\leq \sum_{k=1}^{n-3}\frac{2k+1}{a_1+\ldots+a_k}+\frac{(n-1)^2}{a_1+\ldots+a_{n-2}}. $$

Step5. By Step3, Step4, induction and Step1 again: $$\sum_{k=1}^{n}\frac{2k+1}{a_1+\ldots+a_k}\leq \frac{3}{a_1}+\frac{9}{a_2}+\sum_{j=3}^{n}\frac{4}{a_j}\leq \sum_{j=1}^{n}\frac{4}{a_j}.$$

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In fact, it is much easier to prove the stronger inequality: $$\sum_{k=1}^{n}\frac{2k+1}{a_1+\ldots+a_k}\leq -\frac{n^2}{a_1+\ldots+a_n}+\sum_{k=1}^{n}\frac{4}{a_k}.$$ –  Jack D'Aurizio Oct 30 '12 at 9:17

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