Inequality

Let $a_k>0$. Prove that:

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

Solution



We are splitting the problem into several steps.

1st:  Let $a_1,a_2,\alpha,\beta,\gamma$ be 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 $(\alpha a_2-\beta a_1)^2\geq 0$.

2nd: Let $a_1,a_2,\alpha,\beta,\gamma,\delta$ be 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 Step 1. By induction, it is easy to prove the analogous statement for $k$ variables $a_1,\ldots,a_k$.

3rd: By Step1,
$$\sum_{k=1}^{n}\frac{2k+1}{a_1+\cdots+a_k}-\frac{4}{a_n}\leq \sum_{k=1}^{n-1}\frac{2k+1}{a_1+\cdots+a_k}+\frac{(\sqrt{2n+1}-2)^2}{a_1+\cdots+a_{n-1}}\leq \\
\quad \quad \quad \quad \quad \quad \leq\sum_{k=1}^{n-2}\frac{2k+1}{a_1+\cdots+a_k}+\frac{n^2}{a_1+\cdots+a_{n-1}}$$

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

5th:
By Step3, Step4, induction and Step1 again:
$$\sum_{k=1}^{n}\frac{2k+1}{a_1+\cdots+a_k}\leq \frac{3}{a_1}+\frac{9}{a_2}+\sum_{j=3}^{n}\frac{4}{a_j}\leq \sum_{j=1}^{n}\frac{4}{a_j}$$