pyramid_fan90 wrote: Let $a,b,c>0$ such that $a+b+c=3$. Prove that :$$\frac{ab}{ab+a+b}+\frac{bc}{bc+b+c}+\frac{ca}{ca+c+a}+\frac{1}{9}\left(\frac{(a-b)^2}{ab+a+b}+\frac{(b-c)^2}{bc+b+c}+\frac{(c-a)^2}{ca+c+a}\right)\leq1.$$ The stronger one is more easy. In fact, \[1-{\mspace{-1mu}\left[\sum\frac{ab}{ab+(a+b)\frac{a+b+c}3}+\frac{1}{\color{red}6}\sum\frac{(a-b)^2}{ab+(a+b)\frac{a+b+c}3}\right]\mspace{-1mu}}=\frac{\sum{\!\left(a^4 b^2+a^4 c^2+a^2 b^3 c+a^2 b^3 c+a^2 b c^3+a^2 b c^3+b^3 c^3+b^3 c^3-8a^2b^2c^2\right)}}{2\mspace{-1mu}\prod{\!\left[3ab+(a+b)(a+b+c)\right]}}\ge0\text{.}\]

pyramid_fan90 wrote: Let $a,b,c>0$ such that $a+b+c=3$. Prove that :$$\frac{ab}{ab+a+b}+\frac{bc}{bc+b+c}+\frac{ca}{ca+c+a}+\frac{1}{9}\left(\frac{(a-b)^2}{ab+a+b}+\frac{(b-c)^2}{bc+b+c}+\frac{(c-a)^2}{ca+c+a}\right)\leq1.$$ $$\sum\frac{ab}{ab+a+b}+\frac{1}{9}\sum\frac{(a-b)^2}{ab+a+b}\leq1 \iff \sum \frac{a^2+7ab+b^2}{ab+a+b}\leq 9$$We have $$\frac{a^2+7ab+b^2}{ab+a+b}\leq a+b+1 \iff 0\leq b(a-1)^2+a(b-1)^2$$. So: $$\sum \frac{a^2+7ab+b^2}{ab+a+b}\leq 2(a+b+c)+3=9$$And we are done.

Let $a, b, c$ be positive real numbers such that $a + b + c = 3.$ Prove that the following inequality holds: \begin{center} $\frac{ab}{ab + a + b} + \frac{bc}{bc + b + c} + \frac{ca}{ca + c + a} + \frac{1}{9}\left(\frac{(a - b)^{2}}{ab + a + b} + \frac{(b - c)^{2}}{bc + b + c} + \frac{(a - c)^{2}}{ac + a + c}\right) \leq 1.$

**WLOG we can assume that:** $\frac{a^{2} +7ab+b^{2}}{ab+a+b} \leq ma+nb+k$ $\Longrightarrow m+n+k=3$ $\frac{\partial \left(\frac{a^{2} +7ab+b^{2}}{ab+a+b}\right)}{\partial a}\mid_{a=1} =1=m$, $\frac{\partial \left(\frac{a^{2} +7ab+b^{2}}{ab+a+b}\right)}{\partial b}\mid_{b=1} =1=n$ $\sum_{cyc}\frac{ab}{ab+a+b} +\frac{1}{9}\sum_{cyc}\frac{(a-b)^{2}}{ab+a+b} \leq 1$ $\Longleftrightarrow \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{ab+a+b} \leq \sum_{cyc}(a+b+1)$ $\frac{a^{2} +7ab+b^{2}}{ab+a+b} \leq a+b+1$ $\Longleftrightarrow b(a-1)^{2} +a(b-1)^{2} \geq 0$ $\therefore \frac{a^{2} +7ab+b^{2}}{ab+a+b} \leq a+b+1$ $\Longrightarrow \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{ab+a+b} \leq 9$ $\Longleftrightarrow \sum_{cyc}\frac{ab}{ab+a+b} +\frac{1}{9}\sum_{cyc}\frac{(a-b)^{2}}{ab+a+b} \leq 1.$

$\frac{ab}{ab+a+b} +\frac{bc}{bc+b+c} +\frac{ca}{ca+c+a} +\frac{1}{9}\left(\frac{(a-b)^{2}}{ab+a+b} +\frac{(b-c)^{2}}{bc+b+c} +\frac{(a-c)^{2}}{ac+a+c}\right) \leq 1.$ $\equiv \sum_{cyc}\frac{ab}{ab+a+b} +\frac{1}{9}\sum_{cyc}\frac{(a-b)^{2}}{ab+a+b} \leq 1$ $\equiv \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{9(ab+a+b)} \leq 1$ $\Longleftrightarrow \sum_{cyc}\frac{ab(5-a-b)}{ab+a+b} =\sum_{cyc}\frac{ab(2+c)}{ab+a+b} \leq 3$ $\Longleftrightarrow \frac{1}{9}\sum_{cyc}(2a+2b+c+2+ac+bc) =\frac{1}{9}\left(21+2\sum_{cyc}ab\right) \leq 3$, $\because \left(\sum_{cyc}a\right)^{2} =9\leq \sum_{cyc}ab$ $\therefore 2\sum_{cyc}ab \leq 6$ $\Longrightarrow \frac{1}{9}\left(21+2\sum_{cyc}ab\right) \leq \frac{27}{9} =3.$

$\frac{ab}{ab+a+b} +\frac{bc}{bc+b+c} +\frac{ca}{ca+c+a} +\frac{1}{9}\left(\frac{(a-b)^{2}}{ab+a+b} +\frac{(b-c)^{2}}{bc+b+c} +\frac{(a-c)^{2}}{ac+a+c}\right)$ $\equiv \sum_{cyc}\frac{ab}{ab+a+b} +\frac{1}{9}\sum_{cyc}\frac{(a-b)^{2}}{ab+a+b} \equiv \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{9(ab+a+b)} \equiv \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{3(3a+3b+3ab)}$ $\equiv \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{(a+b+c)(3a+3b) +9ab} \equiv \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{3(a+b)^{2} +9ab+3c(a+b)}$ $\equiv \sum_{cyc}\frac{a^{2} +7ab\ +b^{2}}{3a^{2} +15ab+3b^{2} +3c(a+b)} \Longrightarrow \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{3a^{2} +15ab+3b^{2} +3c(a+b)} -\frac{1}{3}$ $\equiv \sum_{cyc}\frac{6ab-3c(a+b)}{3a^{2} +15ab+3b^{2} +3c(a+b)} \ $ $\therefore \sum_{cyc}\frac{ab}{ab+a+b} +\frac{1}{9}\sum_{cyc}\frac{(a-b)^{2}}{ab+a+b} \leq 1 \Longleftrightarrow \ \sum_{cyc}\frac{2ab-c(a+b)}{ab+a+b} \leq 0$ $\sum_{cyc}\frac{2ab-c(a+b)}{ab+a+b}\overset{\text{C-B-S}}{\leq} \ \frac{1}{9}(12-(\underbrace{\sum\limits_{cyc}\frac{c}{b}}_{\text{\textbf{AM-GM}} \geq 3} +\underbrace{\sum\limits_{cyc}\frac{c}{a}}_{\text{\textbf{AM-GM}} \geq 3} +\underbrace{\sum\limits_{cyc}\left(\frac{bc}{a} +\frac{ca}{b}\right))}_{\text{\textbf{AM-GM}} \geq 2\sum_{cyc} c=6} \leq \frac{1}{9}(12-12) =0.$

### **First - Solution** $\frac{ab}{ab+a+b} +\frac{bc}{bc+b+c} +\frac{ca}{ca+c+a} +\frac{1}{9}\left(\frac{(a-b)^{2}}{ab+a+b} +\frac{(b-c)^{2}}{bc+b+c} +\frac{(a-c)^{2}}{ac+a+c}\right) \leq 1.$ $\equiv \sum_{cyc}\frac{ab}{ab+a+b} +\frac{1}{9}\sum_{cyc}\frac{(a-b)^{2}}{ab+a+b} \leq 1$ $\equiv \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{9(ab+a+b)} \leq 1$ $\Longleftrightarrow \sum_{cyc}\frac{ab(5-a-b)}{ab+a+b} =\sum_{cyc}\frac{ab(2+c)}{ab+a+b} \leq 3$ $\Longleftrightarrow \frac{1}{9}\sum_{cyc}(2a+2b+c+2+ac+bc) =\frac{1}{9}(21+2\sum_{cyc}ab) \leq 3$, because $\left(\sum_{cyc}a\right)^{2} =9\leq \sum_{cyc}ab$ $\therefore 2\sum_{cyc}ab \leq 6$ $\Longrightarrow \frac{1}{9}(21+2\sum_{cyc}ab) \leq \frac{27}{9} =3.$ ### **Second - Solution** $\frac{ab}{ab+a+b} +\frac{bc}{bc+b+c} +\frac{ca}{ca+c+a} +\frac{1}{9}\left(\frac{(a-b)^{2}}{ab+a+b} +\frac{(b-c)^{2}}{bc+b+c} +\frac{(a-c)^{2}}{ac+a+c}\right)$ $\equiv \sum_{cyc}\frac{ab}{ab+a+b} +\frac{1}{9}\sum_{cyc}\frac{(a-b)^{2}}{ab+a+b} \equiv \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{9(ab+a+b)} \equiv \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{3(3a+3b+3ab)}$ $\equiv \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{(a+b+c)(3a+3b) +9ab} \equiv \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{3(a+b)^{2} +9ab+3c(a+b)}$ $\equiv \sum_{cyc}\frac{a^{2} +7ab\ +b^{2}}{3a^{2} +15ab+3b^{2} +3c(a+b)} \Longrightarrow \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{3a^{2} +15ab+3b^{2} +3c(a+b)} -\frac{1}{3}$ $\equiv \sum_{cyc}\frac{6ab-3c(a+b)}{3a^{2} +15ab+3b^{2} +3c(a+b)} \ $ $\therefore \sum_{cyc}\frac{ab}{ab+a+b} +\frac{1}{9}\sum_{cyc}\frac{(a-b)^{2}}{ab+a+b} \leq 1 \Longleftrightarrow \ \sum_{cyc}\frac{2ab-c(a+b)}{ab+a+b} \leq 0$ $\sum_{cyc}\frac{2ab-c(a+b)}{ab+a+b}\overset{\text{C-B-S}}{\leq} \ \frac{1}{9}(12-(\underbrace{\sum\limits_{cyc}\frac{c}{b}}_{\text{\textbf{AM-GM}} \geq 3} +\underbrace{\sum\limits_{cyc}\frac{c}{a}}_{\text{\textbf{AM-GM}} \geq 3} +\underbrace{\sum\limits_{cyc}\left(\frac{bc}{a} +\frac{ca}{b}\right))}_{\text{\textbf{AM-GM}} \geq 2\sum_{cyc} c=6} \leq \frac{1}{9}(12-12) =0.$ ### **Third - Solution** **WLOG we can assume that:** $\frac{a^{2} +7ab+b^{2}}{ab+a+b} \leq ma+nb+k$ $\Longrightarrow m+n+k=3$ $\frac{\partial \left(\frac{a^{2} +7ab+b^{2}}{ab+a+b}\right)}{\partial a}\mid_{a=1} =1=m$, $\frac{\partial \left(\frac{a^{2} +7ab+b^{2}}{ab+a+b}\right)}{\partial b}\mid_{b=1} =1=n$ $\sum_{cyc}\frac{ab}{ab+a+b} +\frac{1}{9}\sum_{cyc}\frac{(a-b)^{2}}{ab+a+b} \leq 1$ $\Longleftrightarrow \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{ab+a+b} \leq \sum_{cyc}(a+b+1)$ $\frac{a^{2} +7ab+b^{2}}{ab+a+b} \leq a+b+1$ $\Longleftrightarrow b(a-1)^{2} +a(b-1)^{2} \geq 0$ $\therefore \frac{a^{2} +7ab+b^{2}}{ab+a+b} \leq a+b+1\Longrightarrow \sum_{cyc}\frac{a^{2} +7ab+b^{2}}{ab+a+b} \leq 9$ $\Longleftrightarrow \sum_{cyc}\frac{ab}{ab+a+b} +\frac{1}{9}\sum_{cyc}\frac{(a-b)^{2}}{ab+a+b} \leq 1.$