Prove the inequality $\sum_{1\le i<j\le n}\sqrt{a_ia_j}<\frac{n-1}2\sum_{i=1}^n a_i$

Question

Please demonstrate this is true This is the exercise: $$\sqrt{a_1a_2}+\sqrt{a_1a_3}+\ldots+\sqrt{a_1a_n}+\sqrt{a_2a_3}+\ldots+\sqrt{a_{n-1}a_n}<\frac{n-1}2(a_1+a_2+a_3+\ldots+a_n).$$ I tried to solve it, but I couldn't do anything right. This is my idea: $\sqrt{a_1a_2}<\frac{a_1+a_2}2$ -because geometric mean < arithmetic mean; $\sqrt{a_1a_3}<\frac{a_1+a_3}2$, $\dots$, $\sqrt{a_{n-1}a_n}<\frac{a_{n-1}+a_n}2$

Please give me the answer!!

$a_1,a_2,a_3,\ldots,a_n$ are real, positive nummbers

Answer 1

Use the AM-GM inequality, which states that:

$$\sqrt{a_ia_j}\le\frac{a_i+a_j}{2}$$

Then we have the following:

$$\sum_{i< j}\sqrt{a_ia_j}\le\sum_{i< j}\frac{a_i+a_j}{2}=\frac{1}{2}\sum_{i< j}a_i+a_j\\=\frac{1}{2}[(a_1+a_2)+(a_1+a_3)+\ldots+(a_1+a_n)+(a_2+a_3)+\ldots+(a_{n-1}+a_n)]$$

Now notice that in the sum on the right hand side, each $a_i$ appears $n-1$ times, so we have:

$$\sum_{i< j}\sqrt{a_ia_j}\le\frac{n-1}{2}(a_1+\ldots+a_n)$$

Answer 2

As $a_i$ s $>0$

$$(\sqrt {a_i}-\sqrt {a_j})^2\ge0\implies a_i+a_j\ge 2\sqrt{a_ia_j}$$

Putting $i=1,2,\cdots,n$ and $j=1,2,\cdots, n$ and adding them we get,

$$(n+1)\sum a_i\ge \sum2\sqrt{a_ia_j}+\sum 2a_i(\text{ this is due to } i=j)$$

$$\implies (n-1)\sum a_i\ge 2\sum\sqrt{a_ia_j}$$

Answer 3

Or

$$ \begin{eqnarray} 0 & \leq &\textrm{variance of the sequence } \sqrt{a_1}, \sqrt{a_2}, \dotsc, \sqrt{a_n} \\ & = &\frac{\sum_k a_k}{n} - \left( \frac{\sum_k \sqrt{a_k}}{n} \right)^2 \\ & = & \frac{n-1}{n^2} \sum_k a_k - \frac{2}{n^2} \sum_{j < k}\sqrt{a_ja_k} \end{eqnarray} $$