If $a_1, a_2, \dots , a_n$ are positive numbers, prove following inequality:

$\sqrt{a_1a_2}+\sqrt{a_2a_3}+ \cdots + \sqrt{a_{n-1}a_n}≤ \frac{n-1}{2}(a_1+a_2+ \cdots + a_n)$

I know the solution, though there may be other algorithms. I will post my solution if there is no identical one.


I assume the left hand side should be

$$\sum_{1 \leq i < j \leq n} \sqrt{a_ia_j}$$

and not

$$\sum_{i = 1}^{n-1}\sqrt{a_ia_{i+1}}$$

since that would be strange and follows from the one I'm gonna show. So I'm gonna prove the following:

$$\sum_{1 \leq i < j \leq n} \sqrt{a_ia_j} \leq \frac{n-1}{2}\sum_{i=1}^na_i$$

This is simple enough since it follows from the AM-GM inequality:

$$\sum_{1 \leq i < j \leq n} \sqrt{a_ia_j} \leq \sum_{1 \leq i < j \leq n}\frac{1}{2}(a_i+a_j)=\frac{n-1}{2}\sum_{i=1}^na_i$$


Arithmetic-Geometric mean inequality says that:

$$\sqrt{a_i a_j} \leq (a_i + a_j)/2 $$

Do this for $(i,j) = (1,2), \dots, (n, n-1)$ and add them all up.

$$ \sqrt{a_1 a_2} + \sqrt{a_2 a_3} + \dots + \sqrt{a_{n-1} a_n} \leq \frac{a_1}{2} + a_2 + ... + a_{n-1} + \frac{a_{n}}{2}$$

For $n > 1$, the result follows.


I Agree with Atticus's answer in that you probably have a typo in the question, for which when fixed as a far tighter inequality.


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