$\Leftrightarrow\sum a\geq{\frac{3}^{2}}+\sum{\frac{ab^2}^{1+b^2}}$ by AM-GM $1+b^2\geq 2b$ and so ${\frac{ab^2}^{2b}}\geq{\frac{ab^2}^{1+b^2}}$ $\sum a\geq{\frac{3}^{2}}+{\sum{\frac{ab}^{2}}}\geq\frac{3}^{2}}+\sum{\frac{ab^2}^{1+b^2}}$ we have to show that $3\geq\sum{ab}$ $\Leftrightarrow{\frac{\sum a}^{3}}\geq{\sqrt{\frac{\sum ab}^{3}}}$ which is the MacLaurin-Inequality

With the same condition ; the stronger result also hold : $ \sum \frac{a}{ \sqrt{b^2 +1}} \ge \frac{3}{ \sqrt{2}}$ And of course ; It's much harder

No one try my problem ????? I 'm a little bit sad Actually ; I think the second ineq is more suitable for such competition like IMO Not very hard ; but nice If nobody give a solution ; I can post it in $ 6^{th}$ November

JohannesG wrote: $\Leftrightarrow\sum a\geq{\frac{3}^{2}}+\sum{\frac{ab^2}^{1+b^2}}$ by AM-GM $1+b^2\geq 2b$ and so ${\frac{ab^2}^{2b}}\geq{\frac{ab^2}^{1+b^2}}$ $\sum a\geq{\frac{3}^{2}}+{\sum{\frac{ab}^{2}}}\geq\frac{3}^{2}}+\sum{\frac{ab^2}^{1+b^2}}$ we have to show that $3\geq\sum{ab}$ $\Leftrightarrow{\frac{\sum a}^{3}}\geq{\sqrt{\frac{\sum ab}^{3}}}$ which is the MacLaurin-Inequality $LHS=\sum{a-\frac{ab^2}{b^2+1}}$

Your proof is wrong If $a+b+c=3$, then $ab^2+bc^2+ca^2\le 3$ does not hold. Take for example $a=1,b=2, c=0$.

Applying AM-GM We'll get : $\dfrac{a}{b^2+1}$=$a-\dfrac{ab^2}{1+b^2}$>$a-\dfrac{ab^2}{2b}$=$a-\dfrac{ab}{2}$ So inequality becomes: $a+b+c-\dfrac{ab+bc+ca}{2}$>$\dfrac{3}{2}$ since $3(ab+bc+ac)$<$(\sum\limits_{cyc})^2 a$=9 This is my proof,Equality holds if a=b=c=1

Here is the solution as I promised ( sorry ; I 'm a little bit busy these days ) It's also the solution my friend gave me in one Math forum in Vietnam : Solution : firstly ; we will introduce the following lemma : lemma : If $a ; b ; c$ are positive real numbers such that : $ a + b + c \le 3$ then : $ (ab + bc + ca ) (a^2 b + b^2 c + c^2 a) \le 9$ ( this lemma is very well - known ; if you cannot prove it ; ask me or ask can_hang2007 for the proof) back to our problem : since : $ 2\sqrt{ 2(b^2 +1)} \le ( b^2 +1 ) + 2 = b^2 +3$ $ \implies \frac{a}{\sqrt{ 2(b^2 +1)} } \ge \frac{2a}{b^2 + 3}$ So ; we only need to prove $ \sum \frac{a}{b^2 + 3} \ge \frac{3}{4} \ \ (*)$ $ \frac{a}{b^2 + 3} = \frac{1}{3} \frac{a(3+b^2) - ab^2}{b^2 +3} = \frac{1}{3} \left( a - \frac{ab^2 }{b^2 + 3} \right)$ So ; it's very easy to see that $(*)$ is equivalent to : $ \sum \frac{ab^2}{b^2 + 3} \le \frac{3}{4}$ Now ; apply AM-GM ineq for four positive real numbers : $ b^2 + 3 = b^2 + 1 +1 +1 \ge 4 \sqrt{b}$ $ \implies \frac{ab^2}{b^2 + 3} \le \frac{ab^2}{4 \sqrt{b}}$ $ \implies \sum \frac{ab^2}{b^2 + 3} \le \sum \frac{ab\sqrt{b} }{4 }$ So ; we need to prove that : $ \sum ab\sqrt{b} \le 3$ $ \iff ( \sum ab\sqrt{b})^2 \le 9 \ \ (**)$ Now ; Applying BCS inequality : $ ( ab \sqrt{b} + bc \sqrt{c} + ca \sqrt{a})^2 \le (ab + bc + ca)(ab^2 + bc^2 + ca^2 ) \le 9$ ( due to the lemma) thus ; $(**)$ is right and the initial ineq is completely proved

nguyenvuthanhha wrote: Here is the solution as I promised ( sorry ; I 'm a little bit busy these days ) It's also the solution my friend gave me in one Math forum in Vietnam : Solution : firstly ; we will introduce the following lemma : lemma : If $a ; b ; c$ are positive real numbers such that : $ a + b + c \le 3$ then : $ (ab + bc + ca ) (a^2 b + b^2 c + c^2 a) \le 9$ ( this lemma is very well - known ; if you cannot prove it ; ask me or ask can_hang2007 for the proof) back to our problem : since : $ 2\sqrt{ 2(b^2 +1)} \le ( b^2 +1 ) + 2 = b^2 +3$ $ \implies \frac{a}{\sqrt{ 2(b^2 +1)} } \ge \frac{2a}{b^2 + 3}$ So ; we only need to prove $ \sum \frac{a}{b^2 + 3} \ge \frac{3}{4} \ \ (*)$ $ \frac{a}{b^2 + 3} = \frac{1}{3} \frac{a(3+b^2) - ab^2}{b^2 +3} = \frac{1}{3} \left( a - \frac{ab^2 }{b^2 + 3} \right)$ So ; it's very easy to see that $(*)$ is equivalent to : $ \sum \frac{ab^2}{b^2 + 3} \le \frac{3}{4}$ Now ; apply AM-GM ineq for four positive real numbers : $ b^2 + 3 = b^2 + 1 +1 +1 \ge 4 \sqrt{b}$ $ \implies \frac{ab^2}{b^2 + 3} \le \frac{ab^2}{4 \sqrt{b}}$ $ \implies \sum \frac{ab^2}{b^2 + 3} \le \sum \frac{ab\sqrt{b} }{4 }$ So ; we need to prove that : $ \sum ab\sqrt{b} \le 3$ $ \iff ( \sum ab\sqrt{b})^2 \le 9 \ \ (**)$ Now ; Applying BCS inequality : $ ( ab \sqrt{b} + bc \sqrt{c} + ca \sqrt{a})^2 \le (ab + bc + ca)(ab^2 + bc^2 + ca^2 ) \le 9$ ( due to the lemma) thus ; $(**)$ is right and the initial ineq is completely proved that's great! nguyenvuthanhha, thank you and I remember the lemma I have proved it in my blog

Very nice proof of the lemma ; Kuing ; Thanks It's much shorter and easier than mine

Plugging (2) and (3) into the RHS of (1), we not get iinequality

the myh123's prove isnot right. who can prove by Cauchy inequality？

who can prove by Cauchy inequality？

\[ \frac{a}{b^{2}+1}+\frac{b}{c^{2}+1}+\frac{c}{a^{2}+1}\geq\frac{3}{2}. \] $\frac{a}{b^2+1}=\frac{a(b^2+1)-ab^2}{b^2+1}=a-\frac{ab^2}{b^2+1}\geq\ a-\frac{ab^2}{2b}=a-\frac{ab}{2}$ $\sum_{cyc} \frac{a}{b^2+1}\geq\ a-\frac{ab}{2}+b-\frac{bc}{2}+c-\frac{ac}{2}=a+b+c-\frac{ab+ac+bc}{2}\geq\ a+b+c-\frac{\frac{(a+b+c)^2}{3}}{2}=3-\frac{a+b+c}{2}=3-\frac{3}{2}=\frac{3}{2}$ it is nice

Uzbekistan wrote: $\ a+b+c-\frac{a^2+b^2+c^2}{2}\geq\ 3-\frac{a+b+c}{2}=3-\frac{3}{2}=\frac{3}{2}$ it is nice It is wrong, try $a,b,c=2,0.5,0.5$ to get $a+b+c-\frac{a^2+b^2+c^2}{2}=2+0.5+0.5-4.5/2<1.5$ so your proof is wrong. To be correct $a+b+c-\frac{ab+bc+ac}{2}\ge a+b+c -\frac{(a+b+c)^2}{6}=1.5$ by $AM-GM$

\[\sum_{cyc}bc \le \frac{1}{3}(\sum_{cyc}a)^2=3\Rightarrow\] \[\sum_{cyc} \frac{a}{b^2+1}=\sum_{cyc}a-\sum_{cyc} \frac{ab^2}{b^2+1}\ge\sum_{cyc}a-\sum_{cyc} \frac{ab}{2}\ge 3-\frac{3}{2}=\frac{3}{2}.\]

Amir Hossein wrote: Let $a, b, c$ be non-negative numbers with $a+b+c = 3$. Prove the inequality \[\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1} \geq \frac 32.\] Let $a, b, c$ be non-negative numbers with $a+b+c = 3$. Prove the inequality $$\frac{a}{ab+1}+\frac{b}{bc+1}+\frac{c}{ca+1}\geq \frac{3}{2}$$

nguyenvuthanhha wrote: With the same condition ; the stronger result also hold : $ \sum \frac{a}{ \sqrt{b^2 +1}} \ge \frac{3}{ \sqrt{2}}$ And of course ; It's much harder one line down

Amir Hossein wrote: Let $a, b, c$ be non-negative numbers with $a+b+c = 3$. Prove the inequality \[\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1} \geq \frac 32.\] Let $a, b, c$ be non-negative numbers with $a+b+c = 3$. Prove the inequality $$\frac a{b^n+1}+\frac b{c^n+1}+\frac c{a^n+1}\geq \frac 32.$$Where $ n\in N^+.$ h

Amir Hossein wrote: Let $a, b, c$ be non-negative numbers with $a+b+c = 3$. Prove the inequality \[\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1} \geq \frac 32.\] Let three non-negative reals $ a,b,c $ satisfy : $ a+b+c=3 $.Prove that$$\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}+\frac{3abc}{2} \le 3$$here