How to show this conjecture：$\frac{1}{\sqrt{a_{1}}}+\frac{1}{\sqrt{a_{2}}}+\cdots+\frac{1}{\sqrt{a_{n}}}\ge\frac{2n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$

Question

I have prove this inequality $$\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\cdots+\dfrac{1}{\sqrt{n}}>2\sqrt{n+1}-2$$ because $$\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}>\sum_{k=1}^{n}\dfrac{2}{\sqrt{k}+\sqrt{k+1}}=2\sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k})=2\sqrt{n+1}-2$$ so $$\sum_{k=1}^{n}\dfrac{1}{\sqrt{k}}>2\sqrt{n+1}-2$$

Now I conjecture：

if postive arithmetic progression $\{a_{n}\}$ ,and the common difference of successive members is $d>0$.have $$\dfrac{1}{\sqrt{a_{1}}}+\dfrac{1}{\sqrt{a_{2}}}+\cdots+\dfrac{1}{\sqrt{a_{n}}}\ge\dfrac{2n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$$

Answer 1

Note that $$ \begin{align} \frac1n\sum_{k=1}^n\frac2{\sqrt{a+kd}+\sqrt{a+(k-1)d}} &=\frac1n\sum_{k=1}^n\frac2d\left(\sqrt{a+kd}-\sqrt{a+(k-1)d}\right)\\ &=\frac2{nd}\left(\sqrt{a+nd}-\sqrt{a}\right)\\[4pt] &=\frac2{\sqrt{a+nd}+\sqrt{a}}\tag{1} \end{align} $$ Let $a_k=\sqrt{a+kd}$. $$ \begin{align} \frac1{n+1}\sum_{k=0}^n\frac1{a_k} &=\frac1{n+1}\left[\frac12\left(\frac1{a_n}+\frac1{a_0}\right)+\sum_{k=1}^n\frac12\left(\frac1{a_k}+\frac1{a_{k-1}}\right)\right]\tag{2}\\ &\ge\frac1{n+1}\left[\frac2{a_n+a_0}+\sum_{k=1}^n\frac2{a_k+a_{k-1}}\right]\tag{3}\\ &=\frac2{a_n+a_0}\tag{4} \end{align} $$ Explanation:
$(2)$: algebra
$(3)$: harmonic mean is less than the arithmetic mean
$(4)$: apply $(1)$

Inequality $(4)$ is the inequality in question.

Answer 2

Using the AM-HM inequality states: if $x_1\leq\ldots\leq x_n$ are nonnegative numbers, then \begin{align} \frac{x_1+\ldots +x_n}{n} \geq \frac{n}{\frac{1}{x_1}+\ldots+\frac{1}{x_n}} \end{align} which implies \begin{align} \frac{1}{x_1} + \ldots + \frac{1}{x_n} \geq \frac{n^2}{x_1+ \ldots + x_n} \geq \frac{n}{x_1 + x_n}. \end{align} However, we might be able to do better.

Special Case: Let us consider arithmetic progression of the form $a_k = dk$ for $k = 1, \dots, n$, then the desired inequality becomes \begin{align} \sum^n_{k=1} \frac{1}{\sqrt{k}} \geq \frac{2n}{1+\sqrt{n}} \ \ \ (*) \end{align} To establish $(\ast)$, we will use Euler Maclaurin Series \begin{align} \sum^n_{k=1} \frac{1}{\sqrt{k}} = 1 + \int^n_1\frac{dx}{\sqrt{x}} + \frac{1}{2}\left(\frac{1-\sqrt{n}}{\sqrt{n}} \right)+\frac{1}{24}\left( 1-n^{-3/2}\right)+R \end{align} where \begin{align} |R| \leq \frac{3\zeta(2)}{2(2\pi)^2}\int^n_1 \frac{dx}{x^{5/2}} = \frac{1}{24}(1-n^{-3/2}). \end{align} Then it follows \begin{align} \sum^n_{k=1} \frac{1}{\sqrt{k}}-\frac{2n}{1+\sqrt{n}} =&\ 1+ \frac{2(n-1)}{1+\sqrt{n}}+\frac{1}{2}\left(\frac{1-\sqrt{n}}{\sqrt{n}} \right)-\frac{2n}{1+\sqrt{n}}+\text{ something positive}\\ \geq&\ \frac{\sqrt{n}-1}{\sqrt{n}+1}+\frac{1}{2}\left(\frac{1-\sqrt{n}}{\sqrt{n}} \right) = \frac{1}{2}\frac{(\sqrt{n}-1)^2}{\sqrt{n}(\sqrt{n}+1)}.\\ \end{align}

Thus, it follows \begin{align} \sum^n_{k=1} \frac{1}{\sqrt{k}} \geq \frac{2n}{1+\sqrt{n}}. \end{align}

Comment: Using the same idea, we might be able to work with arithmetic progression of the form $a_k = d(m+k)$ for $k=1, \ldots, n$. However, I have no clue how to get the most general arithmetic progression $a_k = d(m+k)+\ell$.

Answer 3

Using the same method, I can show the following slightly weaker result.

Let $a_i = a_0 + id$, for $1 \le i \le n+1$. Since $d > 0$, we have $a_{i+1} > a_i$. Hence $$ \frac{1}{\sqrt{a_i}} > \frac{2}{\sqrt{a_{i+1}} + \sqrt{a_i}} = \frac{2(\sqrt{a_{i+1}} - \sqrt{a_i})}{a_{i+1} - a_i} = \frac{2(\sqrt{a_{i+1}} - \sqrt{a_i})}{d}. $$

This gives $$ \sum_{i=1}^n \frac{1}{\sqrt{a_i}} > \frac{2}{d} \sum_{i=1}^n \left( \sqrt{a_{i+1}} - \sqrt{a_i} \right) = \frac{2(\sqrt{a_{n+1}} - \sqrt{a_1})}{d}.$$

Finally, we simplify, getting $$ \sum_{i=1}^n \frac{1}{\sqrt{a_i}} > \frac{2(a_{n+1} - a_1)}{d(\sqrt{a_{n+1}} + \sqrt{a_1})} = \frac{2n}{\sqrt{a_1} + \sqrt{a_{n+1}}}.$$