Prove that if $a,b,c>0$ and $a+b+c=1,$ then $$\frac{bc+a+1}{a^2+1}+\frac{ca+b+1}{b^2+1}+\frac{ab+c+1}{c^2+1}\leq \frac{39}{10}$$
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Tags: inequalities, JBMO, romania
31.05.2015 01:24
For #1 Because of $\frac{bc+a+1}{a^2+1}\le{\frac{\frac{(b+c)^2}{4}+a+1}{a^2+1}}=\frac{\frac{(1-a)^2}{4}+a+1}{a^2+1}\le{\frac{34}{25}-\frac{9a}{50}}$
06.06.2015 13:31
$$\frac{bc+a+1}{a^2+1}+\frac{ca+b+1}{b^2+1}+\frac{ab+c+1}{c^2+1}\leq \frac{39}{10}$$ If we substract 3 from both side we get, $\sum{\frac{bc+a(1-a)}{a^2+1}=\sum {\frac{bc+ac+ab}{a^2+1}}}\leq \frac {9}{10}$
24.07.2015 06:36
Let $a,b,c>0$ and $a+b+c=1.$ Prove that $$(ab+bc+ca)(\frac{1}{9a^2+1}+\frac{1}{9b^2+1}+\frac{1}{9c^2+1})\leq \frac{1}{2}.$$
28.08.2015 03:55
Prove that if $a,b,c>0$ and $a+b+c=1,$ then $$ \frac{a^2+bc}{a^2+1}+\frac{b^2+ca}{b^2+1}+\frac{c^2+ab}{c^2+1}\le\frac{13}{20}.$$
19.01.2016 05:32
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!
11.03.2016 22:21
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??
12.03.2016 00:28
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 $$\frac{bc+a+1}{a^2+1}+\frac{ca+b+1}{b^2+1}+\frac{ab+c+1}{c^2+1}\geq \frac{13}{10(a^2+b^2+c^2)}$$
12.03.2016 03:23
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 $$ \frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\leq \frac{27}{10}.$$
12.03.2016 05:37
sqing wrote: Let $a,b,c>0$ and $a+b+c=1.$ Prove that $$(ab+bc+ca)(\frac{1}{9a^2+1}+\frac{1}{9b^2+1}+\frac{1}{9c^2+1})\leq \frac{1}{2}.$$
12.03.2016 11:41
$\frac{\frac{(1-a)^2}{4}+a+1}{a^2+1}\le{\frac{34}{25}-\frac{9a}{50}}$ 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 $a = \frac{1}{2}$.
13.03.2016 01:28
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.$$ \Rightarrow \sum {\frac{1}{{1 + {a^2}}}} \le \frac{{27}}{{10}}$ It’s true, use problem on and inequality $\sum {ab \le \frac{1}{3}} {\left( {\sum a } \right)^2} = \frac{1}{3}.\,\,\,$.
03.11.2016 12:44
Excuse me, Nguyenngoctu, I think that your solution is very good, but did you try to find -27/50a+27/25 randomly, or there existed the logic? Thanks in advance.
25.05.2018 16:22
Here's my solution to the original problem
Meanwhile, here's the one for squing's first problem
Also, Squing's 4th and 5th problem are trivialized by Jensen's
07.07.2019 15:49
socrates wrote: Prove that if $a,b,c>0$ and $a+b+c=1,$ then $$\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}$$ 2015 Junior Balkan Team Selection Tests-Romania
07.07.2019 16:20
MilosMilicev wrote: Excuse me, Nguyenngoctu, I think that your solution is very good, but did you try to find -27/50a+27/25 randomly, or there existed the logic? Thanks in advance. Tangent Line Method...
24.03.2024 16:39
Let $ a,b,c $ be real numbers such that $ a+b+c=1.$ Prove that$$ \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