Inequality proof $\frac{n}{a_1a_{2n+1}}<\frac{1}{a_1a_2}+\frac{1}{a_3a_4}+…+\frac{1}{a_{2n-1}a_{2n}}<\frac{n}{a_0a_{2n}}$

Question

Can someone give me an idea about how to prove the inequality? Mention that $a_0, a_1, a_2, ..., a_{2n}, a_{2n+1}$ is an AP and $0<a_0<a_1<...<a_{2n}<a_{2n+1}$ $$\frac{n}{a_1a_{2n+1}}<\frac{1}{a_1a_2}+\frac{1}{a_3a_4}+...+\frac{1}{a_{2n-1}a_{2n}}<\frac{n}{a_0a_{2n}}$$ Mention that I tried ideas like: $$\frac{1}{a_{2n+1}}<\frac{1}{a_1}<\frac{1}{a_{0}};$$ and then add, but with no result. A possible solution is: $$\frac{n}{a_1a_{2n+1}}<\frac{1}{r}(\frac{1}{a_1}-\frac{1}{a_2}+\frac{1}{a_3}-\frac{1}{a_4}+...+\frac{1}{a_{2n-1}}-\frac{1}{a_{2n}})<\frac{n}{a_0a_{2n}}, r=a_n-a_{n-1} $$

$$\frac{rn}{a_1a_{2n+1}}<\frac{1}{a_1}+\frac{1}{a_3}+...+\frac{1}{a_{2n-1}}-(\frac{1}{a_2}+\frac{1}{a_4}+...+\frac{1}{a_{2n}})<\frac{rn}{a_0a_{2n}} $$ Let be $S_1>I_1$: $$\frac{1}{a_1}+\frac{1}{a_1}+...+\frac{1}{a_1}>\frac{1}{a_1}+\frac{1}{a_3}+...+\frac{1}{a_{2n-1}} $$ Let be $S_2>I_2$: $$\frac{1}{a_2}+\frac{1}{a_2}+...+\frac{1}{a_2}>\frac{1}{a_2}+\frac{1}{a_4}+...+\frac{1}{a_{2n}} $$ The following difference can be taken into account in this case? $$S_1-S_2>I_1-I_2 \rightarrow S_1-S_2=\frac{nr}{a_1a_2} $$ Then $$ \frac{nr}{a_1a_{2n+1}}<\frac{nr}{a_1a_2}<\frac{nr}{a_0a_{2n}} $$ Inequality comes out immediately if this difference occurs. Can give me some advice to solve it or if this approach is wrong, why is it wrong? Thank you!

Answer 1

For simplicity let us assume that the common difference of the arithmetic progression is equal to one (the same calculation works in the general case with small modifications).

You correctly noticed that the middle expression can be written as $$ \frac{1}{a_1a_2}+\frac{1}{a_3a_4}+\cdots+\frac{1}{a_{2n-1}a_{2n}} \\ = \frac{1}{a_1}-\frac{1}{a_2} + \frac{1}{a_3}-\frac{1}{a_4} + \cdots + \frac{1}{a_{2n-1}} - \frac{1}{a_{2n}} \, . $$ Now the “trick“ is to group this sum as $$ = \frac{1}{2a_0} - \left( \frac{1}{2a_0}-\frac{1}{a_1}+\frac{1}{2a_2}\right) - \cdots - \left( \frac{1}{2a_{2n-2}}-\frac{1}{a_{2n-1}}+\frac{1}{2a_{2n}}\right) - \frac{1}{2a_{2n}} \, . $$ and show that all terms in the parentheses are positive. It follows that the whole expression is $$ < \frac{1}{2a_0}- \frac{1}{2a_{2n}} = \frac{n}{a_0 a_{2n}} $$ which proves the right inequality.

The proof of the left inequality is similar.