<s>[INST] Proving an identity involving terms in arithmetic progression. If $a_1,\ldots,a_n$ are in arithmetic progression and $a_i\gt 0$ for all $i$, then how to prove the following two identities:

$ (1)\large  \frac{1}{\sqrt{a_1} + \sqrt{a_2}} + \frac{1}{\sqrt{a_2} + \sqrt{a_3}} + \cdots + \frac{1}{\sqrt{a_{n-1}} + \sqrt{a_n}} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n}}$
$(2) \large\frac{1}{a_1 \cdot a_n} + \frac{1}{a_2 \cdot a_{n-1}} + \frac{1}{a_3 \cdot a_{n-2}}+ \cdots + \frac{1}{a_n \cdot a_1} = \frac{2}{a_1 + a_n} \biggl( \frac{1}{a_1} + \frac{1}{a_2} + \cdots +  \frac{1}{a_n} \biggr)$

 [/INST] Both identities can be proved quite easily with inductions. Let $d$ be the common difference, i.e. $d=a_{n+1}-a_n$.


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*Use induction. So we need to prove
$$ \frac{n-2}{\sqrt{a_1}+\sqrt{a_ {n-1}}}+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_ n}}=\frac{n-1}{\sqrt{a_1}+\sqrt{a_ n}}.$$
Rationalize the denominators and substitute $$a_{n-1}-a_1=(n-2)d, a_n-a_{n-1}=d, a_n-a_1=(n-1)d.$$ Upon cancellations, we find
$$\sqrt{a_{n-1} }-\sqrt{a_ 1}+\sqrt{a_n }-\sqrt{a_{n-1} }=\sqrt{a_ n}-\sqrt{a_ 1}$$ on the left and the same on the right.

*Use induction, assuming true for $m<n$. Canceling the sum $$ \frac{1}{a_1a_n}+\frac{1}{a_na_1}$$ from the left with the term $$\frac{2}{a_1+a_n}\left(\frac{1}{a_1}+\frac{1}{a_n}\right)$$ on the right and using the induction hypothesis on the remaining terms $$\frac{1}{a_2a_{n-1}}+\cdots +\frac{1}{a_{n-1}a_2},$$ we find that  we need to prove  $$\frac{2}{a_2+a_{n-1}}$$ on the left is equal to $$\frac{2}{a_1+a_{n}}$$ on the right. And this is true because $a_2+a_{n-1}=a_1+a_n=2a_1+(n-1)d$. 
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