Multiply both sides by $ 2(a_{1}+a_{2}+...+a_{n})=2S$ and see that: $ \sum \frac{2a_{i}a_{j}S}{a_{i}+a_{j}}= \sum \frac{2a_{i}a_{j}(a_{i}+a_{j})}{a_{i}+a_{j}}+\sum \frac{2a_{i}a_{i}(S-a_{i}-a_{j})}{a_{i}+a_{j}}= 2 \sum a_{i}a_{j}+\sum \frac{2a_{i}a_{i}(S-a_{i}-a_{j})}{a_{i}+a_{j}}$. So we want to prove that: $ \sum \frac{2a_{i}a_{i}(S-a_{i}-a_{j})}{a_{i}+a_{j}}\leq (n-2) \sum a_{i}a_{j}$. But for $ i \not = j \not = k \not = i$ we have $ \frac{2a_{i}a_{j}a_{k}}{a_{i}+a_{j}}\leq \frac{1}{2}(a_{i}a_{k}+a_{j}a_{k})$ since we can divide by $ a_{k}$ and transform it to $ (a_{i}-a_{j})^{2}\geq 0$. So after writing such inequalities and summing them up we get the desired inequality.

The inequality is equivalent to $ \sum_{i < j}\frac{(a_{i}-a_{j})^{2}}{a_{i}+a_{j}}\geq \frac{\sum_{i < j}(a_{i}-a_{j})^{2}}{\sum a_{i}}$ and the result follows...

RDeepMath91 > Could you show me more precisly how did you get this inequality's form please "And the result follows" : please, explain more, I don't see how.

The QuattoMaster 6000 wrote:

I think the first point has to be changed and how did you prove this second one?

Use the identity $\dfrac {2a_ia_j} {a_i + a_j} = a_i + a_j - \dfrac {a_i^2 + a_j^2} {a_i + a_j}$ to obtain \begin{align*} \sum a_ia_j &\ge \dfrac {\sum a_i} {n} \sum \dfrac {2a_ia_j} {a_i + a_j} \iff\\ \sum a_ia_j &\ge \dfrac {\sum a_i} {n} \left (\sum a_i + a_j - \sum \dfrac {a_i^2 + a_j^2} {a_i + a_j} \right) \iff\\ \sum a_ia_j + \dfrac {\sum a_i} {n}\sum \dfrac {a_i^2 + a_j^2} {a_i + a_j} &\ge \dfrac {\sum a_i} {n} (n - 1) (\sum a_i) \iff \\ \dfrac {\sum a_i} {n} \sum \dfrac {a_i^2 + a_j^2} {a_i + a_j} & \ge \dfrac {n - 1} {n} (\sum a_i^2 + 2\sum a_i a_j) - \sum a_i a_j \iff \\ \sum a_i \sum \dfrac {a_i^2 + a_j^2} {a_i + a_j} & \ge (n - 1) \sum a_i^2 + (n - 2) \sum a_i a_j \iff \\ (n - 1)\sum a_i^2 + \sum (\sum_{k \ne i, j} a_k) \dfrac {a_i^2 + a_j^2} {a_i + a_j} &\ge (n - 1) \sum a_i^2 + (n - 2) \sum a_i a_j \iff \\ \sum (\sum_{k \ne i, j} a_k) \dfrac {a_i^2 + a_j^2} {a_i + a_j} & \ge (n - 2) \sum a_i a_j \end{align*} Since AM-GM gives us $\dfrac {a_i^2 + a_j^2} {a_i + a_j} \ge \dfrac {a_i + a_j} {2}$, we have \begin{align*} \sum (\sum_{k \ne i, j} a_k) \dfrac {a_i^2 + a_j^2} {a_i + a_j} \ge \sum (\sum_{k \ne i, j} a_k) \dfrac {a_i + a_j} {2} = (n - 2) \sum a_i a_j \end{align*}as desired.

The inequality is equivalent with: $ \sum_{i < j}{\frac{a_{i}a_{j}}{a_{i}+a_{j}}}((n-1)(a_{1}+a_{2}+\cdots+a_{n}))\leq\frac{n(n-1)}{2}\cdot\sum_{i < j}{a_{i}a_{j}}$. Let $x_{ij}=\frac{a_{i}a_{j}}{a_{i}+a_{j}}$ and $y_{ij}=a_i+a_j$. Now we apply Chebyshev's inequality to numbers $x_{ij}$ and $y_{ij}$ and we will get inequality on the beginning.

TheBernuli wrote: The inequality is equivalent with: $ \sum_{i < j}{\frac{a_{i}a_{j}}{a_{i}+a_{j}}}((n-1)(a_{1}+a_{2}+\cdots+a_{n}))\leq\frac{n(n-1)}{2}\cdot\sum_{i < j}{a_{i}a_{j}}$. Let $x_{ij}=\frac{a_{i}a_{j}}{a_{i}+a_{j}}$ and $y_{ij}=a_i+a_j$. Now we apply Chebyshev's inequality to numbers $x_{ij}$ and $y_{ij}$ and we will get inequality on the beginning. I had the same idea but can you please elaborate how you ordered the sequences ? I guess the ordering of x_{ij}'s and y_{ij} 's can not be uniquely ordered if we ONLY assume a certain ordering of variables a_1,...a_n . So , in that case , in case of some variables a_1,a_2,...,a_n , Can you please show that x_{ij} and y_{ij} are similarly sorted. ? Plz elaborate it if possible

My solution: Let $x_{ij}=a_ia_j$ and $y_{ij}=a_i+a_j$ and $S=\sum a_i$ \begin{align*} RHS-LHS&=\sum_{i<j}\frac{nx_{ij}}{2S}- \frac{x_{ij}}{y_{ij}} \\ &=\sum_{i<j} x_{ij} . \frac{\sum_{k=1}^n a_i-a_k+a_j-a_k}{2.S.y_{ij}} \\ &=\sum_{i<k} (a_i-a_k).\sum_{j\not\in \lbrace i,k \rbrace }( \frac{x_{ij}}{2S.y_{ij}}-\frac{x_{jk}}{2Sy_{jk}}) \\ &= \sum_{i<k} \frac{a_i-a_k}{2.S}.\sum_{j\not\in \lbrace i,k \rbrace } a_j(\frac{a_i}{y_{ij}}-\frac{a_k}{y_{jk}}) \\ &= \sum_{i<k} \frac{(a_i-a_k)^2}{2.S}.\sum_{j\not\in \lbrace i,k \rbrace } a_j^2 \geq 0 \end{align*} QED .$ \blacksquare$

Hello: My solution: We will use induction on $n$.For $n=2$ the inequality is true (as an equality).Suppose that for $n=k$ the inequality is also true. Now,consider a $k+1$ positive reals $a_1,a_2,...,a_k$.Since the inequality is symmetric,we can suppose that $a_1\leq a_2\leq \ldots a_{k+1}$. We have to show that $\left(\sum_{1\leq i<j\leq k+1} a_ia_j\right)\cdot \frac{k+1}{2\sum_{i=1}^{k+1} a_i}\geq \sum_{1\leq i<j\leq k+1} \frac{a_ia_j}{a_i+a_j} \ (\bigstar)$. However,we have the following equalities: $\bullet$ $\sum_{1\leq i<j\leq k+1} a_ia_j=\left(\sum_{1\leq i<j\leq k} a_ia_j\right)+a_{k+1}\sum_{i=1}^{k} a_i$. $\bullet$ $\sum_{1\leq i<j\leq k+1} \frac{a_ia_j}{a_i+a_j}=\left(\sum_{1\leq i<j\leq k} \frac{a_ia_j}{a_i+a_j}\right)+a_{k+1}\sum_{i=1}^{k} \frac{a_i}{a_i+a_{k+1}}$. Thus,$(\bigstar)$ becomes $\left[\left(\sum_{1\leq i<j\leq k} a_ia_j\right)+a_{k+1}\sum_{i=1}^{k} a_i\right]\cdot \frac{k+1}{2\sum_{i=1}^{k+1} a_i}\geq \left(\sum_{1\leq i<j\leq k} \frac{a_ia_j}{a_i+a_j}\right)+a_{k+1}\sum_{i=1}^{k} \frac{a_i}{a_i+a_{k+1}} \ (1)$ We know (from our inductive hypothesis) that $\sum_{1\leq i<j\leq k} \frac{a_ia_j}{a_i+a_j}\leq \frac{k}{2\sum_{i=1}^{k} a_i}\cdot \left(\sum_{1\leq i<j\leq k} a_ia_j\right)$, thus,instead of $(1)$,it suffices to show that $\left[\left(\sum_{1\leq i<j\leq k} a_ia_j\right)+a_{k+1}\sum_{i=1}^{k} a_i\right]\cdot \frac{k+1}{2\sum_{i=1}^{k+1} a_i}\geq \frac{k}{2\sum_{i=1}^{k} a_i}\cdot \left(\sum_{1\leq i<j\leq k} a_ia_j\right)+a_{k+1}\sum_{i=1} \frac{a_i}{a_i+a_{k+1}}$. The function $f(x)=\frac{x}{x+a_{k+1}}$ is concave,thus,from Jensen we obtain $\sum_{i=1}^{k} \frac{a_i}{a_i+a_{k+1}}\leq \frac{k\sum_{i=1}^{k}a_i}{ka_{k+1}+\sum_{i=1}^{k} a_i}$.Thus,it suffices to show that $\left[\left(\sum_{1\leq i<j\leq k} a_ia_j\right)+a_{k+1}\sum_{i=1}^{k} a_i\right]\cdot \frac{k+1}{2\sum_{i=1}^{k+1} a_i}\geq \frac{k}{2\sum_{i=1}^{k} a_i}\cdot \left(\sum_{1\leq i<j\leq k} a_ia_j\right)+a_{k+1}\frac{k\sum_{i=1}^{k}a_i}{ka_{k+1}+\sum_{i=1}^{k} a_i}$ which is equivalent to $\frac{1}{2}\left(\sum_{1\leq i<j\leq k} a_ia_j\right)\left(\frac{k+1}{\sum_{i=1}^{k+1} a_i}-\frac{k}{\sum_{i=1}^{k} a_i}\right)\geq a_{k+1}\sum_{i=1}^{k} a_i\left(\frac{k}{ka_{k+1}+\sum_{i=1}^{k} a_i}-\frac{(k+1)}{2\sum_{i=1}^{k+1} a_i}\right)$ $\Leftrightarrow \frac{1}{2}\left(\sum_{1\leq i<j\leq k} a_ia_j\right)\frac{\sum_{i=1}^{k}a_i-ka_{k+1}}{\left(\sum_{i=1}^{k+1}a_i\right)\left(\sum_{i=1}^{k} a_i\right)}\geq a_{k+1}(k-1)\sum_{i=1}^{k}a_i\cdot \frac{\sum_{i=1}^{k} a_i-ka_{k+1}}{2\left(\sum_{i=1}^{k+1}a_i\right)\left(ka_{k+1}+\sum_{i=1}^{k} a_i\right)}$. Since $a_{k+1}=\max \bigcup_{i=1}^{k+1} a_i$ we get $ka_{k+1}\geq \sum_{i=1}^{k} a_i$ thus,the latter inequality is equivalent to $\frac{\left(\sum_{1\leq i<j\leq k} a_ia_j\right)}{\sum_{i=1}^{k}a_i}\leq \frac{a_{k+1}(k-1)\sum_{i=1}^{k} a_i}{ka_{k+1}+\sum_{i=1}^{k} a_i}\Leftrightarrow \sum_{1\leq i<j\leq k}a_ia_j\leq \frac{(k-1)\left(\sum_{i=1}^{k} a_i\right)^2}{k+\frac{1}{a_{k+1}}\cdot \sum_{i=1}^{k} a_i}$. Since $a_{k+1}=\max \bigcup_{i=1}^{k+1} a_i$ we get $\frac{\sum_{i=1}^{k}a_i}{a_{k+1}}\leq k$ thus $\frac{(k-1)\left(\sum_{i=1}^{k} a_i\right)^2}{k+\frac{1}{a_{k+1}}\cdot \sum_{i=1}^{k} a_i}\geq \frac{(k-1)\left(\sum_{i=1}^{k} a_i\right)^2}{2k}$. Hence,it suffices to show that $\left(\sum_{i=1}^{k}a_i\right)^2\geq \frac{2k}{k-1}\sum_{1\leq i<j\leq k} a_ia_j\Leftrightarrow \sum_{i=1}^k a_i^2\geq \frac{2}{k-1}\sum_{1\leq i<j\leq k} a_ia_j$. However,from AM-GM we get $\sum_{1\leq i<j\leq k} a_i^2+a_j^2\geq 2\sum_{1\leq i<j\leq k} a_ia_j$.Each $a_i$ appears in $k-1$ pairs,thus $LHS=(k-1)\sum_{i=1}^{k} a_i^2$ thus $(k-1)\sum_{i=1}^{k} a_i^2\geq 2\sum_{1\leq i<j\leq k}a_ia_j$ q.e.d.

Wrong solution.

Modesti wrote: By Titu's inequality and Cauchy-Schwarz, $$\sum_{i<j}\dfrac{a_ia_j}{a_i+a_j}\le \dfrac{(\sum_{i<j}\sqrt{a_ia_j})^2}{2(a_1+a_2+...+a_n)}\le\dfrac{n}{2(a_1+a_2+...+a_n)}\sum_{i<j}a_ia_j$$ and we are done. $\square$ Check please your first step.

You are right, I reversed the sign. Thank you. Will delete the wrong solution.

We will prove the inequality with both sides multiplied by $\sum_{i=1}^n a_i:$ $$\sum_{i<j} \left( \frac{a_ia_j}{a_i+a_j} \right) \cdot \sum_{i=1}^n a_i=\sum_{i<j} a_ia_j +\sum_{i<j} \frac{a_ia_j \sum_{k\neq i,j} a_k}{a_i+a_j} \stackrel{\text{AM-HM}}{\leq} \sum_{i<j} a_ia_j+\sum_{i<j} \frac{(a_i+a_j) \sum_{k\neq i,j} a_k}{4}=$$$$=\sum_{i<j} a_ia_j+\frac{(n-2)}{2}\sum_{i<j}a_ia_j=\frac n2 \sum_{i<j} a_ia_j\quad \blacksquare$$

Let $S=a_1+a_2+\dots + a_n$. For the sake of brevity, all summations will denote either the sum from $1$ to $n$, and with $i<j$ if there are two variables. Note that \[(a_i+a_j)-4\left(\frac{a_ia_j}{a_i+a_j}\right)=\frac{(a_i+a_j)^2-4a_ia_j}{a_i+a_j}=\frac{(a_i-a_j)^2}{a_i+a_j}\]Thus, \[(n-1)S-4\sum{\frac {a_{i}a_{j}}{a_{i} + a_{j}}}=\sum{\frac{(a_i-a_j)^2}{a_i+a_j}}\]Meanwhile, \begin{align*} (n-1)S-4\left(\frac {n}{2S}\cdot \sum{a_{i}a_{j}}\right) &= \frac{(n-1)S^2-2n\sum{a_ia_j}}{S} \\ &= \frac{(n-1)^2\sum{a_i^2}+2(n-2)\sum{a_ia_j}-2n\sum{a_ia_j}}{S} \\ &= \frac{(n-1)^2\sum{a_i^2}-2\sum{a_ia_j}}{S} \\ &= \frac{\sum{(a_i-a_j)^2}}{S} \end{align*}Therefore, it suffices to show the following inequality, which is obvious because $a_i+a_j \le S$. \[\sum{\frac{(a_i-a_j)^2}{a_i+a_j}} \ge \frac{\sum{(a_i-a_j)^2}}{S}\]

Just an exercise in summations, but took me 2 hours $$\sum_{i<j}\frac{a_ia_j}{a_i+a_j}\cdot\sum_{k=1}^{n}a_k\le\sum_{i<j}a_ia_j+\sum_{i<j} \frac{a_ia_j\sum_{k\neq i,j}a_k}{a_i+a_j}\stackrel{\text{AM-HM}}{\le}\sum_{i<j}a_ia_j+\sum_{i<j} \frac{(a_i+a_j)\sum_{k\neq i,j} a_k}{4}=\frac{n}{2}\sum_{i<j}a_ia_j$$$$\iff\sum_{i<j}\frac{(a_i+a_j)\sum_{k\neq i,j}a_k}{2}=(n-2)\sum_{i<j}a_ia_j=(n-2)n.$$But this is evident, since $a_i,a_k$ (and $a_j,a_k$) satisfy the same conditions as $i<j$, with any $a_la_m$ appearing n-2 times when $i=l, j\ne l,m$ or when $i=m, j\ne l,m$. (Note that even if $i=l>j\ne l,m$ or the same with m, we can just switch i and j since there's only one way they can be ordered increasing.)

the more I practice inequalities, the more I don't like them...