<s>[INST] How to show this conjecture：$\frac{1}{\sqrt{a_{1}}}+\frac{1}{\sqrt{a_{2}}}+\cdots+\frac{1}{\sqrt{a_{n}}}\ge\frac{2n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$ 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
  $$\dfrac{1}{\sqrt{a_{1}}}+\dfrac{1}{\sqrt{a_{2}}}+\cdots+\dfrac{1}{\sqrt{a_{n}}}\ge\dfrac{2n}{\sqrt{a_{1}}+\sqrt{a_{n}}}$$

 [/INST] 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
$$ \sum_{i=1}^n \frac{1}{\sqrt{a_i}} > \frac{2(a_{n+1} - a_1)}{d(\sqrt{a_{n+1}} + \sqrt{a_1})} = \frac{2n}{\sqrt{a_1} + \sqrt{a_{n+1}}}.$$
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