The first one is very easy: 1/c(a+b)+1/b(a+c)+1/a(b+c) \geq 27/2(a+b+c) ² 9/2(ab+bc+ca) \geq 27/2(a+b+c) ² 1/(ab+bc+ca) \geq 3/(a+b+c) ² (a+b+c) ² \geq 3(ab+bc+ca) which is obviously true Cheers

Rewrite the inequality like: \[\left[\sum_{cyc} (a+b)\right]\left[\sum_{cyc} \frac1{b(a+b)}\right]\left[\sum_{cyc} a\right] \ge 27\] and apply AM -GM to the three sums and the inequality is solved!!

First inequality is : 1/b(a+b) + 1/c(b+c) + 1/a(c+a) \geq ....... and not 1/a(b+c) + 1/b(a+c) +1/c(a+b) \geq ........ I think it is a different inequality.

hmm.. but my method still works, doesn't it?

Lagrangia, you are right and your proof is very neat but I referred to Maverik's proof. My proof is: changing a with b the inequality remains valid and is transformed in (1/a(b+a)) +(1/c(a+c))+(1/b(b+c)) \geq ...... summing the two inequalities we obtain (a+b+c)/abc \geq 27/(a+b+c)^2 whic can be rewritten as (a+b+c)^3 \geq 27abc true by AM-GM inequality.

Yeah sorry about that! But using rearangement inequality we get: 1/b(a+b)+1/c(b+c)+1/a(a+c) \geq 1/c(a+b)+1/a(b+c)+1/b(a+c) so then we use \sum 1/c(a+b) \geq 9/2(ab+bc+ca) ........ and so on.....

Another solution is : From AM-GM ineq we have $\frac{2}{b(a+b)}+\frac{2}{c(b+c)}+\frac{2}{a(c+a)}\geq\frac{6}{XY}$ where $X=\sqrt[3]{abc}$ and $Y=\sqrt[3]{(a+b)(b+c)(c+a)}$ By AM-GM again we have that $X\leq\frac{a+b+c}{3}$ and $Y\leq\frac{2a+2b+2c}{3}$ The result follows

Yes, AM-Gm twice, Cauchy, rearangement, or direct calculations work on this inequality. Anyway, this inequality has been proposed by Greece, not by L. Panaitopol .

Yes exactly, but I was afraid to tell this. But now I am sure. The proposer is Sotiris Louridas. Additionally the best solution is the one which use Caushy in Engel form. I anyone want it i will post it

The Cauchy solution is well-known. Actually, I have found a nice solution using rearangement inequality and then, I reduced it to Nesbitt.

Could you post it ? thanks. But with details. With no use of trivial etc

Well, by rearangement we have \[ \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\geq \frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}. \] Actually we will prove this stronger inequality: \[ \frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}\geq\frac{9}{2(ab+bc+ca)}. \] Well, this las inequality is equivalent with \[ \frac{ab+bc+ca}{a(b+c)}+\frac{ab+bc+ca}{b(c+a)}+\frac{ab+bc+ca}{c(a+b)}\geq\frac{9}{2}. \] This last inequality is \[ \frac{bc}{a(b+c)}+\frac{ca}{b(c+a)}+\frac{ab}{c(a+b)}\geq\frac{3}{2}. \] Well, by making the substitutions $bc=m, ca=n$ and $ca=p$ the inequality finally reduces to \[ \frac{m}{n+p}+\frac{n}{p+m}+\frac{p}{m+n}\geq\frac{3}{2}. \] From here you can apply Nesbitt to finish.

I hope this solution is in detail.

cezar lupu wrote: I hope this solution is in detail. Maybe the only of yours

cezar lupu wrote: Well, by rearangement we have \[ \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\geq \frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}. \] Actually we will prove this stronger inequality: \[ \frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}\geq\frac{9}{2(ab+bc+ca)}. \] Well, this las inequality is equivalent with \[ \frac{ab+bc+ca}{a(b+c)}+\frac{ab+bc+ca}{b(c+a)}+\frac{ab+bc+ca}{c(a+b)}\geq\frac{9}{2}. \] This last inequality is \[ \frac{bc}{a(b+c)}+\frac{ca}{b(c+a)}+\frac{ab}{c(a+b)}\geq\frac{3}{2}. \] Well, by making the substitutions $bc=m, ca=n$ and $ca=p$ the inequality finally reduces to \[ \frac{m}{n+p}+\frac{n}{p+m}+\frac{p}{m+n}\geq\frac{3}{2}. \] From here you know how to proceed. You are perfect.I admire you very much.

Anyway, who is perfect?

razec upul wrote: Anyway, who is perfect? I don't know, but I invented something . If you write the name razec upul from the other side it is Cesar Lupu . And if you see the flag in Cesar Lupu and razec upul then obviously you conclude that Cesar Lupu and razec upul are the same.

Very nice Silouan. That was hard.

Why you did this ?

starcraft wrote: cezar lupu wrote: Well, by rearangement we have \[ \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\geq \frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}. \] Actually we will prove this stronger inequality: \[ \frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}\geq\frac{9}{2(ab+bc+ca)}. \] Well, this las inequality is equivalent with \[ \frac{ab+bc+ca}{a(b+c)}+\frac{ab+bc+ca}{b(c+a)}+\frac{ab+bc+ca}{c(a+b)}\geq\frac{9}{2}. \]This last inequality is \[ \frac{bc}{a(b+c)}+\frac{ca}{b(c+a)}+\frac{ab}{c(a+b)}\geq\frac{3}{2}. \] Well, by making the substitutions $bc=m, ca=n$ and $ca=p$ the inequality finally reduces to \[ \frac{m}{n+p}+\frac{n}{p+m}+\frac{p}{m+n}\geq\frac{3}{2}. \] From here you know how to proceed. You are perfect.I admire you very much. Nice

manlio wrote: Prove that for all positive real numbers $a,b,c$ the following inequality takes place \[ \frac{1}{b(a+b)}+ \frac{1}{c(b+c)}+ \frac{1}{a(c+a)} \geq \frac{27}{2(a+b+c)^2} . \]Laurentiu Panaitopol, Romania

By Rearrangement we have \[ \frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}\geq \frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)} \]also by Cauchy-Schwarz and AM-GM we have \[ 2(a+b+c)^2(\frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}) \geq 6(ab+ac+bc)(\frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)})=3[a(b+c)+b(c+a)+c(a+b)](\frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)}) \geq 3(1+1+1)^2=27 \]then $$\frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)} \geq \frac{27}{2(a+b+c)^2}$$

sqing wrote: manlio wrote: Prove that for all positive real numbers $a,b,c$ the following inequality takes place \[ \frac{1}{b(a+b)}+ \frac{1}{c(b+c)}+ \frac{1}{a(c+a)} \geq \frac{27}{2(a+b+c)^2} . \]Laurentiu Panaitopol, Romania we will prove this stronger inequality $\frac{1}{b(a+b)}+ \frac{1}{c(b+c)}+ \frac{1}{a(c+a)} \geq\frac{9}{2(ab+bc+ca)}$ $\iff\frac{2(ab+bc+ca)}{b(a+b)}+\frac{2(ab+bc+ca)}{c(b+c)}+\frac{2(ab+bc+ca)}{a(c+a)}\geq 9.$ By AM-GM, $\sum\frac{2(ab+bc+ca)}{b(a+b)}= \sum\frac{a+c}{a+b}+\sum\frac{c}{b}+\sum\frac{a(b+c)}{b(a+b)}\geq3+3+3=9.$ 2009 Belarus: Prove that any positive real numbers a,b,c satisfy the inequlaity$$\frac{1}{(a+b)b}+\frac{1}{(b+c)c}+\frac{1}{(c+a)a}\ge \frac{9}{2(ab+bc+ca)}$$

Here is my solution i know there above correct solution also but i m just sending this for just store it and someone might understand this if something wrong in this please let me know My solution: We can see that here is very standard Problem So Lets do it Lets Apply AM-GM here, \[ \frac{1}{b(a+b)}+ \frac{1}{c(b+c)}+ \frac{1}{a(c+a)} \geq \frac{3}{\sqrt[3]{abc((a+b)(b+c)(c+a))}}\]So to make it Simple Let $X=\sqrt[3]{abc}$ and $Y=\sqrt[3]{(a+b)(b+c)(c+a)}$ So now its looks like \[ \frac{1}{b(a+b)}+ \frac{1}{c(b+c)}+ \frac{1}{a(c+a)} \geq \frac{3}{XY}\]now its looks nice then Apply some AM-GM Lets see what we find \[3\sqrt[3]{abc}=3X\leq a+b+c\implies X\leq \frac{a+b+c}{3}\]Similarly \[3\sqrt[3]{(a+b)(b+c)(c+a)}=3Y\leq a+b+b+c+c+a=2(a+b+c)\implies Y\leq \frac{2(a+b+c)}{3}\]so we can clearly see that \[XY\leq \frac{2(a+b+c)^2}{9}\]\[\implies\frac{1}{XY}\geq \frac{9}{2(a+b+c)^2}\]Now we have to Just plug it Up and we are done Demon Killed. \[ \frac{1}{b(a+b)}+ \frac{1}{c(b+c)}+ \frac{1}{a(c+a)} \geq \frac{3}{\sqrt[3]{abc((a+b)(b+c)(c+a))}}\geq \frac{3}{XY}\geq \frac{3.9}{2(a+b+c)^2}\]\[\implies \frac{1}{b(a+b)}+ \frac{1}{c(b+c)}+ \frac{1}{a(c+a)}\geq \frac{27}{2(a+b+c)^2}\]And we are done demon killed.Mission Accomplished.

Nice one@above

sqing wrote:

That's a easy one. Apply Titu's lemma and done.

@above Titu doesn't work here directly. sqing wrote: 2009 Belarus: Prove that any positive real numbers a,b,c satisfy the inequlaity$$\frac{1}{(a+b)b}+\frac{1}{(b+c)c}+\frac{1}{(c+a)a}\ge \frac{9}{2(ab+bc+ca)}$$ Note that it's equivalent to $$2\sum_{\textsf{cyc}}{a^{4}b^{2}} + 2\sum_{\textsf{cyc}}{a^{4}bc}+2\sum_{\textsf{cyc}}{a^{3}b^{3}}\ge a^{3}b^{2}c+b^{3}c^{2}a+c^{3}a^{2}b+3\sum_{\textsf{cyc}}{a^{3}bc^{2}}+6a^{2}b^{2}c^{2}$$But \begin{align*} 2\sum_{\textsf{cyc}}{a^{4}b^{2}}\ge 2\sum_{\textsf{cyc}}{a^{3}bc^{2}} \hspace{2 mm} \textsf{and} \hspace{2 mm} (4, 1, 1)\ge (2, 2, 2) \end{align*}So it remains to show $$(3, 3, 0)\ge (3, 2, 1)\hspace{2 mm} \textsf{which is true by Muirhead.}$$

Many unnecessarily lengthy solutions and how is “Holder” never mentioned in the thread - posting for strorage. $$(\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(a+c)})(b+c+a)((a+b)+(b+c)+(a+c)) \geq 27$$by Holder’s Inequality and the conclusion follows readily.

Instant Holder... \begin{align*} \left(\frac{1}{b(a+b)}+ \frac{1}{c(b+c)}+ \frac{1}{a(c+a)}\right) \cdot (a+b+c)\cdot (a+b+b+c+c+a)\geq 3^3=27. \end{align*}

manlio wrote: Prove that for all positive real numbers $a,b,c$ the following inequality takes place \[ \frac{1}{b(a+b)}+ \frac{1}{c(b+c)}+ \frac{1}{a(c+a)} \geq \frac{27}{2(a+b+c)^2} . \]Laurentiu Panaitopol, Romania please correct the proposer (as well as the title) to: Sotiris Louridas, Greece (see posts #09,10)

Let $a, b, c$ be positive real numbers. Prove that $$\dfrac{1}{b(a + 2b)} + \frac{1}{c(b +2 c)}+ \frac{1}{a(c + 2a)} \ge \frac{9}{(a + b + c)^2}$$$$\dfrac{1}{b(a +3b)} + \frac{1}{c(b +3c)}+ \frac{1}{a(c +3a)} \ge \frac{27}{4(a + b + c)^2}$$

Let $a, b, c$ be positive real numbers. Prove that $$\dfrac{1}{b(a +kb)} + \frac{1}{c(b +kc)}+ \frac{1}{a(c +ka)}\ge \frac{27}{(k+1)(a + b + c)^2}$$Where $k>0.$

Generalization 1 Let $a_{1}, a_{2},\cdots,a_{2k+1}$ be positive reels. Then prove that $$\sum_{n=1}^{2k+1}{\dfrac{1}{a_{n}\left(a_{n-2k+1}+a_{n-2k+2}+\cdots+a_{n}\right)}}\geq \dfrac{(2k+1)^3}{2k\left(\sum_{cyc}{a_{k}}\right)}$$