For #1 Because of $\frac{bc+a+1}{a^2+1}\le{\frac{\frac{(b+c)^2}{4}+a+1}{a^2+1}}=\frac{\frac{(1-a)^2}{4}+a+1}{a^2+1}\le{\frac{34}{25}-\frac{9a}{50}}$

$$\frac{bc+a+1}{a^2+1}+\frac{ca+b+1}{b^2+1}+\frac{ab+c+1}{c^2+1}\leq \frac{39}{10}$$ If we substract 3 from both side we get, $\sum{\frac{bc+a(1-a)}{a^2+1}=\sum {\frac{bc+ac+ab}{a^2+1}}}\leq \frac {9}{10}$

Let $a,b,c>0$ and $a+b+c=1.$ Prove that $$(ab+bc+ca)(\frac{1}{9a^2+1}+\frac{1}{9b^2+1}+\frac{1}{9c^2+1})\leq \frac{1}{2}.$$

Prove that if $a,b,c>0$ and $a+b+c=1,$ then $$ \frac{a^2+bc}{a^2+1}+\frac{b^2+ca}{b^2+1}+\frac{c^2+ab}{c^2+1}\le\frac{13}{20}.$$

sqing wrote: Let $a,b,c>0$ and $a+b+c=1.$ Prove that $$(ab+bc+ca)(\frac{1}{9a^2+1}+\frac{1}{9b^2+1}+\frac{1}{9c^2+1})\leq \frac{1}{2}.$$ Hi sqing, could you give a proof of this? Thanks in advance!

IstekOlympiadTeam wrote: $$\frac{bc+a+1}{a^2+1}+\frac{ca+b+1}{b^2+1}+\frac{ab+c+1}{c^2+1}\leq \frac{39}{10}$$ If we substract 3 from both side we get, $\sum{\frac{bc+a(1-a)}{a^2+1}=\sum {\frac{bc+ac+ab}{a^2+1}}}\leq \frac {9}{10}$ Can you more elobrate your solution??

socrates wrote: Prove that if $a,b,c>0$ and $a+b+c=1,$ then $$\frac{bc+a+1}{a^2+1}+\frac{ca+b+1}{b^2+1}+\frac{ab+c+1}{c^2+1}\leq \frac{39}{10}$$ The following inequality is also true. Prove that if $a,b,c>0$ and $a+b+c=1,$ then $$\frac{bc+a+1}{a^2+1}+\frac{ca+b+1}{b^2+1}+\frac{ab+c+1}{c^2+1}\geq \frac{13}{10(a^2+b^2+c^2)}$$

Let $a,b,c\geq0$ such that $a+b+c=1$ . Prove that: $$\frac{bc}{a^2+1}+\frac{ca}{b^2+1}+\frac{ab}{c^2+1}\leq\frac{3}{10}$$Let $a,b,c\ge -\frac{3}{4}$ and $a+b+c=1$ . Prove that $$\frac{a}{a^2+1}+\frac{b}{b^2+1}+\frac{c}{c^2+1} \le \frac{9}{10}$$Let $a,b,c>0$ and $a+b+c=1$. Prove that $$ \frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\leq \frac{27}{10}.$$

sqing wrote: Let $a,b,c>0$ and $a+b+c=1.$ Prove that $$(ab+bc+ca)(\frac{1}{9a^2+1}+\frac{1}{9b^2+1}+\frac{1}{9c^2+1})\leq \frac{1}{2}.$$

$\frac{\frac{(1-a)^2}{4}+a+1}{a^2+1}\le{\frac{34}{25}-\frac{9a}{50}}$ wrong with $a = \frac{1}{2}$. \[1.3 = \frac{{\frac{1}{4}{{\left( {1 - a} \right)}^2} + a + 1}}{{{a^2} + 1}} > 1.27 = - \frac{9}{{50}}a + \frac{{34}}{{25}}\]with $a = \frac{1}{2}$.

Prove that if $a,b,c>0$ and $a+b+c=1,$ then: $$\frac{bc+a+1}{a^2+1}+\frac{ca+b+1}{b^2+1}+\frac{ab+c+1}{c^2+1}\leq \frac{39}{10}$$Solution by tuchuyenhg. we have $\begin{array}{l} \sum {\frac{{bc + a + 1}}{{{a^2} + 1}}} \le \frac{{39}}{{10}} \Leftrightarrow \sum {\left( {\frac{{bc + a + 1}}{{{a^2} + 1}} - 1} \right)} \le \frac{9}{{10}} \Leftrightarrow \sum {\frac{{bc + a(1 - a)}}{{{a^2} + 1}}} \le \frac{9}{{10}}\\ \Leftrightarrow \sum {ab} .\sum {\frac{1}{{{a^2} + 1}}} \le \frac{9}{{10}} \end{array}$ hence $\frac{1}{{1 + {a^2}}} \le - \frac{{27}}{{50}}a + \frac{{27}}{{25}} \Leftrightarrow \left( {4 - 3a} \right){\left( {3a - 1} \right)^2} \ge 0,\,\,\forall a \le 1.$$ \Rightarrow \sum {\frac{1}{{1 + {a^2}}}} \le \frac{{27}}{{10}}$ It’s true, use problem on and inequality $\sum {ab \le \frac{1}{3}} {\left( {\sum a } \right)^2} = \frac{1}{3}.\,\,\,$.

Excuse me, Nguyenngoctu, I think that your solution is very good, but did you try to find -27/50a+27/25 randomly, or there existed the logic? Thanks in advance.

Here's my solution to the original problem

Meanwhile, here's the one for squing's first problem

Also, Squing's 4th and 5th problem are trivialized by Jensen's

socrates wrote: Prove that if $a,b,c>0$ and $a+b+c=1,$ then $$\frac{bc+a+1}{a^2+1}+\frac{ca+b+1}{b^2+1}+\frac{ab+c+1}{c^2+1}\leq \frac{39}{10}$$ 2015 Junior Balkan Team Selection Tests-Romania

MilosMilicev wrote: Excuse me, Nguyenngoctu, I think that your solution is very good, but did you try to find -27/50a+27/25 randomly, or there existed the logic? Thanks in advance. Tangent Line Method...

Let $ a,b,c $ be real numbers such that $ a+b+c=1.$ Prove that$$ \dfrac{bc}{ a^2 + 1 } + \dfrac{ca}{ b^2 + 1 } + \dfrac{ab}{ c^2 + 1 } \leq \frac {3}{10}$$$$ \dfrac{a}{ a^2 + 1 } + \dfrac{b}{ b^2 + 1 } + \dfrac{c}{ c^2 + 1 } \leq \frac {9}{10}$$$$ \dfrac{1}{ a^2 + 1 } + \dfrac{1}{ b^2 + 1 } + \dfrac{1}{ c^2 + 1 } \leq \frac {27}{10}$$ sqing wrote: socrates wrote: Prove that if $a,b,c>0$ and $a+b+c=1,$ then $$\frac{bc+a+1}{a^2+1}+\frac{ca+b+1}{b^2+1}+\frac{ab+c+1}{c^2+1}\leq \frac{39}{10}$$ 2015 Junior Balkan Team Selection Tests-Romania