$\sum{\frac{a}{a^2+2}}\leq\sum{\frac{a}{2a+1}}$. We shall prove that $\sum{\frac{a}{2a+1}}\leq{1}$ or $\sum{\frac{2a}{2a+1}}\leq{2}$ or $\sum{\frac{1}{2a+1}}\geq{1}$. Clearing the denominator we have to prove that: $2\sum{a}\geq{6}$ wich is true by AM-GM.

a/a^2+2=<a/2a+1

XAKKER wrote: a/a^2+2=<a/2a+1 right, because of : $ a^2+2 = a^2+1+1 \geq 2a +1$ ( by ineq AM-GM)

n0vakovic wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Proove that \[ \frac a{a^{2} + 2} + \frac b{b^{2} + 2} + \frac c{c^{2} + 2}\leq 1 \] It was post here http://www.mathlinks.ro/viewtopic.php?t=172711&start=20

It was a problem of Baltic way 2005.

n0vakovic wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Proove that \[ \frac a{a^{2} + 2} + \frac b{b^{2} + 2} + \frac c{c^{2} + 2}\leq 1 \] The following inequality is also holds: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Proove that \[ \frac a{a^{2} + 3} + \frac b{b^{2} + 3} + \frac c{c^{2} +3}\leq \frac{3}{4} \] I think it is more difficult than above inequality!!!

We have $ \sum{\frac{a}{a^2 + 2}} = \sum{\frac{a}{a^2 + 1^2 + 1}} \leq \sum{\frac{a}{2a + 1}}$, because $ a^2 + 1 \geq 2a$ Now its enough to prove that $ \sum{\frac{a}{2a + 1}} \leq 1 \leftrightarrow \frac{1}{2}\sum{\frac{2a}{2a + 1}} \leq 1 \leftrightarrow \frac{1}{2}\sum{\frac{2a + 1 - 1}{2a + 1}} \leq 1 \leftrightarrow \frac{1}{2}(1 + 1 + 1 - \sum{\frac{1}{2a + 1}}) \leq 1 \leftrightarrow (3 - \sum{\frac{1}{2a + 1}}) \leq 2 \leftrightarrow \sum{\frac{1}{2a + 1}}) \geq 3 - 2 = 1 \leftrightarrow \sum{\frac{bc}{2abc + bc}}) \geq 1 \leftrightarrow \sum{\frac{\sqrt{bc}^2}{2 + bc}}) \geq 1$ From the Andreescu Inequality we have: $ \sum{\frac{{\sqrt{bc}}^2}{2 + bc}} \geq \frac{{(\sqrt{ba} + \sqrt{bc} + \sqrt{ca})}^2}{6 + bc + ca + ba} = K$ So its enough to prove that $ K \geq 1 \leftrightarrow {(\sqrt{ba} + \sqrt{bc} + \sqrt{ca})}^2 \geq 6 + bc + ca + ba \leftrightarrow ba + bc + ca + 2\sum_{cyclic}{a\sqrt{bc}} \geq 6 + bc + ca + ba \leftrightarrow 2\sum_{cyclic}{a\sqrt{bc}} \geq 6 \leftrightarrow \sum_{cyclic}{a\sqrt{bc}} \geq 3$ Which is true because from the AM-GM Inequality we have: ${{ \sum_{cyclic}{a\sqrt{bc}} \geq 3\sqrt[3]{abc\sqrt{ba}\sqrt{ca}\sqrt{cb}} = 3\sqrt[3]{abc\sqrt{a^2b^2c^2}} = 3\sqrt[3]{abcabc}} = 3\sqrt[3]{1}} = 3$

zaizai-hoang wrote: The following inequality is also holds: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Proove that \[ \frac a{a^{2} + 3} + \frac b{b^{2} + 3} + \frac c{c^{2} + 3}\leq \frac {3}{4} \] Your inequality can be proved by the $ \sum_{cyc}\left(\frac{1}{4}-\frac{a}{a^2+3}+\frac{\ln a}{8}\right)\geq0$ idea. For the following \[ \frac a{a^{2} + 4} + \frac b{b^{2} + 4} + \frac c{c^{2} + 4}\leq \frac {3}{5}\] this idea doesn't help but the inequality seems true. By the way, $ \frac a{a^{2} + 5} + \frac b{b^{2} + 5} + \frac c{c^{2} + 5}\leq \frac {1}{2}$ is wrong.

arqady wrote: zaizai-hoang wrote: The following inequality is also holds: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Proove that \[ \frac a{a^{2} + 3} + \frac b{b^{2} + 3} + \frac c{c^{2} + 3}\leq \frac {3}{4} \] Your inequality can be proved by the $ \sum_{cyc}\left(\frac {1}{4} - \frac {a}{a^2 + 3} + \frac {\ln a}{8}\right)\geq0$ idea. For the following \[ \frac a{a^{2} + 4} + \frac b{b^{2} + 4} + \frac c{c^{2} + 4}\leq \frac {3}{5} \] this idea doesn't help but the inequality seems true. By the way, $ \frac a{a^{2} + 5} + \frac b{b^{2} + 5} + \frac c{c^{2} + 5}\leq \frac {1}{2}$ is wrong. But \[ \sum_{cyc}\left(\frac {1}{4} - \frac {a}{a^2 + 3} + \frac {\ln a}{8}\right)\geq0\] is not correct for all $ a\ge 0$ Anyway, it is not hard to prove

Sailor wrote: $ \sum{\frac {2a}{2a + 1}}\leq{2}$ or $ \sum{\frac {1}{2a + 1}}\geq{1}$. How does the first imply the second?

$ \displaystyle \frac {2a}{2a+1} = \frac {(2a+1) - 1}{2a+1} = 1 - \frac {1}{2a+1}$. Do the same for $ b$ and $ c$ and then add them.

zaizai-hoang wrote: n0vakovic wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Proove that \[ \frac a{a^{2} + 2} + \frac b{b^{2} + 2} + \frac c{c^{2} + 2}\leq 1 \] The following inequality is also holds: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Proove that \[ \frac a{a^{2} + 3} + \frac b{b^{2} + 3} + \frac c{c^{2} + 3}\leq \frac {3}{4} \] I think it is more difficult than above inequality!!! It here http://www.mathlinks.ro/viewtopic.php?p=1098909#1098909

for the first one $ \sum \frac{a}{a^2+2}\leq\sum \frac{a}{2a+1}=\sum \frac{1}{2+bc}$ now we have to prove $ \sum \frac{1}{2+bc}\leq1$ equivalent to $ a+b+c\geq3$,which is true

arqady wrote: By the way, $ \frac a{a^{2} + 5} + \frac b{b^{2} + 5} + \frac c{c^{2} + 5}\leq \frac {1}{2}$ is wrong. It is true, arqady.

Vasc wrote: arqady wrote: By the way, $ \frac a{a^{2} + 5} + \frac b{b^{2} + 5} + \frac c{c^{2} + 5}\leq \frac {1}{2}$ is wrong. It is true, arqady. For $ abc=1$ or for all $ a,b$ and $ c$?

For any positive real numbers such that $ abc=1$, of course.

anyway, what is andreescu inequality?

meganyot wrote: anyway, what is andreescu inequality? $ \frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\cdots+\frac{a_n^2}{b_n}\ge\frac{(a_1+a_2+\ldots+a_n)^2}{b_1+b_2+\ldots+b_n}$

Vasc wrote: arqady wrote: By the way, $ \frac a{a^{2} + 5} + \frac b{b^{2} + 5} + \frac c{c^{2} + 5}\leq \frac {1}{2}$ is wrong. It is true, arqady. It seems you are right, but I still can't prove it.

This is my proof.

Thank you very much, Vasc!

Vasc wrote: arqady wrote: By the way, $ \frac a{a^{2} + 5} + \frac b{b^{2} + 5} + \frac c{c^{2} + 5}\leq \frac {1}{2}$ is wrong. It is true, arqady. I found now a proof by Cauchy Schwarz Inequality for this hard inequality!

Can you post it?

n0vakovic wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Proove that \[ \frac a{a^{2} + 2} + \frac b{b^{2} + 2} + \frac c{c^{2} + 2}\leq 1\] $ \frac a{a^{2} + k} + \frac b{b^{2} + k} + \frac c{c^{2} + k}\leq {\frac{3}{1+k}}$ holds $ \forall {a,b,c}>0,abc=1\iff{k_1\le{k}\le{k_2}}$ in which $ k_1={\frac{{59-9\sqrt{59}-3{\sqrt{646-118{\sqrt{29}}}}}}{4}}=0 .19725809...,$ $ k_2={\frac{{59-9\sqrt{59}+3{\sqrt{646-118{\sqrt{29}}}}}}{4}}=5.06950026....$

can_hang2007 wrote: Vasc wrote: arqady wrote: By the way, $ \frac a{a^{2} + 5} + \frac b{b^{2} + 5} + \frac c{c^{2} + 5}\leq \frac {1}{2}$ is wrong. It is true, arqady. I found now a proof by Cauchy Schwarz Inequality for this hard inequality! Congratulation can_hang2007! I really want to see your Cauchy-Schwarz proof. Could you post it?

Does this way work for this problem: $ \frac{4a}{a^2+5} \le \frac{x+1}{x^2+x+1}$ with $ x=\frac{1}{a^{\frac{3}{4}}}$

speciallove wrote: Does this way work for this problem: $ \frac {4a}{a^2 + 5} \le \frac {x + 1}{x^2 + x + 1}$ with $ x = \frac {1}{a^{\frac {3}{4}}}$ Your way is a very interesting, but this inequality is not hold, try $ a=2,$ this inequality become $ 8x^{2}\le x+1,$ where $ x=\frac{1}{2^{\frac{3}{4}}}<1$ $ \Rightarrow2\sqrt{2}=8x^{2}\le x+1<2$ contradiction.

sorry,I want to say $ k= \frac{-4}{3}$ not $ \frac{-3}{4}$ Because I will find k for the inequality such that: $ \frac{4a}{a^2+5} \le \frac{a^k+1}{a^{2k}+a^k+1}$ or $ f(k)=(a^k+1)(a^2+5)-4a(a^{2k}+a^k+1) \ge 0$ f('1)=6k+8=0[/tex] or $ k= \frac{-4}{3}$ with $ k=\frac{-4}{3}$ we have $ x=\frac{1}{a^{\frac{4}{3}}}$

speciallove wrote: sorry,I want to say $ k = \frac { - 4}{3}$ not $ \frac { - 3}{4}$ Because I will find k for the inequality such that: $ \frac {4a}{a^2 + 5} \le \frac {a^k + 1}{a^{2k} + a^k + 1}$ or $ f(k) = (a^k + 1)(a^2 + 5) - 4a(a^{2k} + a^k + 1) \ge 0$ f('1)=6k+8=0[/tex] or $ k = \frac { - 4}{3}$ with $ k = \frac { - 4}{3}$ we have $ x = \frac {1}{a^{\frac {4}{3}}}$ But this inequality is still not hold for all $ a>0,$ try $ a=\frac{1}{4},$ this inequality become $ 16x^{2}\le65x+65,$ where $ x=4^{\frac{4}{3}}>6$ $ \Rightarrow 65x+65<65x+11x=76x<16x^{2}$ contradiction.

Vasc wrote: arqady wrote: By the way, $ \frac a{a^{2} + 5} + \frac b{b^{2} + 5} + \frac c{c^{2} + 5}\leq \frac {1}{2}$ is wrong. It is true, arqady. This is just... so easy Here is a pretty elementary proof.

Very nice proof, Inequalities Master

fjwxcsl wrote: n0vakovic wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Proove that \[ \frac a{a^{2} + 2} + \frac b{b^{2} + 2} + \frac c{c^{2} + 2}\leq 1\] $ \frac a{a^{2} + k} + \frac b{b^{2} + k} + \frac c{c^{2} + k}\leq {\frac{3}{1+k}}$ holds $ \forall {a,b,c}>0,abc=1\iff{k_1\le{k}\le{k_2}}$ in which $ k_1={\frac{{59-9\sqrt{59}-3{\sqrt{646-118{\sqrt{29}}}}}}{4}}=0 .19725809...,$ $ k_2={\frac{{59-9\sqrt{59}+3{\sqrt{646-118{\sqrt{29}}}}}}{4}}=5.06950026....$ But can we use a way to get the best k ?

fjwxcsl wrote: n0vakovic wrote: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Proove that \[ \frac a{a^{2} + 2} + \frac b{b^{2} + 2} + \frac c{c^{2} + 2}\leq 1\] $ \frac a{a^{2} + k} + \frac b{b^{2} + k} + \frac c{c^{2} + k}\leq {\frac{3}{1+k}}$ holds $ \forall {a,b,c}>0,abc=1\iff{k_1\le{k}\le{k_2}}$ in which $ k_1={\frac{{59-9\sqrt{59}-3{\sqrt{646-118{\sqrt{29}}}}}}{4}}=0 .19725809...,$ $ k_2={\frac{{59-9\sqrt{59}+3{\sqrt{646-118{\sqrt{29}}}}}}{4}}=5.06950026....$ Let me ask you this: Why?

n0vakovic wrote: Let $a$, $b$, $c$ be positive real numbers such that $abc=1$. Prove that \[\frac a{a^{2}+2}+\frac b{b^{2}+2}+\frac c{c^{2}+2}\leq 1 \] Let $a=\dfrac{x}{y}, b=\dfrac{y}{z}, =\dfrac{z}{x}$ $\sum{\frac{a}{a^2+2}}=\sum{\dfrac{\dfrac{x}{y}}{\dfrac{x^2}{y^2}+2}=\sum{\dfrac{xy}{x^2+2y^2}}\leq\sum{\dfrac{xy}{2xy+y^2}}=\sum{\dfrac{x}{2x+y}}}$ $\sum{\dfrac{x}{2x+y}}\leq1$ $\Longleftrightarrow\sum{\dfrac{2x}{2x+y}}\leq 2$ $\Longleftrightarrow\sum{-\dfrac{2x}{2x+y}}\geq-2$ $\Longleftrightarrow\sum{1-\dfrac{2x}{2x+y}}\geq1$ $\Longleftrightarrow\sum{\dfrac{y}{2x+y}}\geq1$ $\Longleftrightarrow \dfrac{y}{2x+y}+\dfrac{z}{2y+z}+\dfrac{x}{2z+x}\geq1$ $\Longleftrightarrow \dfrac{y^2}{2xy+y^2}+\dfrac{z^2}{2yz+z^2}+\dfrac{x^2}{2xz+x^2}\geq\dfrac{(x+y+z)^2} {x^2+y^2+z^2+2xy+2xz+2yz}=1$ and we are done.

for #1 With condition $abc=1$,We have \[1-\sum{\frac{a}{a^2+2}}=\frac{\sum{(9+16c^2)(a-b)^2}+\sum{(23+8c+20c^2)(a+b-2)^2}}{36(a^2+2)(b^2+2)(c^2+2)}\] Notice: The condition $abc=1$ can be replaced by $a+b+c=1(or a^2+b^2+c^2=1 or ab+bc+ca=1)$, inequality still holds!

zaizai-hoang wrote: The following inequality is also holds: Let $ a$, $ b$, $ c$ be positive real numbers such that $ abc = 1$. Proove that \[ \frac a{a^{2} + 3} + \frac b{b^{2} + 3} + \frac c{c^{2} +3}\leq \frac{3}{4} \] I think it is more difficult than above inequality!!! See here: http://www.artofproblemsolving.com/community/c6h1195493p5852610

solution here:

Let $a$, $b$, $c$ be positive real numbers such that $ab+bc+ca=1$. Prove that $$ \frac{a}{a^{2}+2}+\frac{b}{b^{2}+2}+\frac{c}{c^{2}+2}\leq \frac{3\sqrt 3}{7}$$