Darij, what happened to you? Or is it Lagrangia hidden?

I don't get it!

harazi wrote: Darij, what happened to you? Or is it Lagrangia hidden? I don't get it either... I just wanted to have all Baltic Way problems posted on MathLinks with solutions... dg

Don't you guys see any similarities between your posts? I mean, the technique of posting the statement of the problem and then the solution was initiated by Lagrangia, I remember. This was the reason for which I asked the question. Sorry, there was no bad intention, just a stupid joke.

Here is my solution: The first thing that is clear is that it's enough to prove the inequality $\frac{1}{a+b+1} + \frac{1}{b+c+1} + \frac{1}{c+a+1} \leq 1$ for any three positive numbers a, b, c with abc = 1; then, by applying this to the numbers $a=p^n$, $b=q^n$, $c=r^n$ (note that pqr = 1 clearly implies $p^n q^n r^n = 1$), we will get the inequality $\frac{1}{p^n+q^n+1} + \frac{1}{q^n+r^n+1} + \frac{1}{r^n+p^n+1} \leq 1$ for every n. Proofs of $\frac{1}{a+b+1} + \frac{1}{b+c+1} + \frac{1}{c+a+1} \leq 1$ can be found in the thread http://www.mathlinks.ro/Forum/viewtopic.php?t=19469

let $p=x^{\frac{3}{n}},q=y^{\frac{3}{n}},r=z^{\frac{3}{n}}$ then$xyz=1$ it suffices to prove$\sum\frac{1}{x^3+y^3+xyz}=1$ by Muirhead,LHS$\le\sum\frac{1}{x^2y+xy62+xyz}=\sum\frac{1}{xy(x+y+z)}=1$ QED

These inequalities holds also for all positive real numbers $a,b,c$ satisfying $a^2+b^2+c^2=3$: $(a)\ \ \ \frac a{b+c+2}+ \frac b{c+a+2}+ \frac c{a+b+2}\ge \frac3{4};$ $(b)\ \ \ \frac a{b+c+3}+ \frac b{c+a+3}+ \frac c{a+b+3}\le \frac3{5}.$

If we put $p^n=a^2, q^n=b^2, r^n=c^2$, it's enough to prove $\sum{\dfrac{1}{a^2+b^2+1}}\leq1$ where $abc=1$ $\sum{\dfrac{1}{a^2+b^2+1}}\leq1$ $\Longleftrightarrow \sum{-\dfrac{1}{a^2+b^2+1}}\geq-1$ $\Longleftrightarrow \sum{1-\dfrac{1}{a^2+b^2+1}}\geq2$ $\Longleftrightarrow \sum{\dfrac{a^2+b^2}{a^2+b^2+1}}\geq2$ From Cauchy-Schwarz, we have $\sum{\dfrac{a^2+b^2}{a^2+b^2+1}}\geq \dfrac{\left(\sum{\sqrt{a^2+b^2}}\right)^2}{2(a^2+b^2+c^2)+3}=\dfrac{2(a^2+b^2+c^2)+2\sum{\sqrt{a^2+b^2}\sqrt{a^2+c^2}}}{2(a^2+b^2+c^2)+3}$ So we're done if the last term is $\geq2$ From another Cauchy Schwarz, we have $\sqrt{a^2+b^2}\sqrt{a^2+c^2}\geq a^2+bc$ So, $\dfrac{2(a^2+b^2+c^2)+2\sum{\sqrt{a^2+b^2}\sqrt{a^2+c^2}}}{2(a^2+b^2+c^2)+3}\geq\dfrac{2(a^2+b^2+c^2)+2(a^2+b^2+c^2+ab+ac+bc)}{2(a^2+b^2+c^2)+3}$ $=\dfrac{4(a^2+b^2+c^2)+2(ab+bc+ac)}{2(a^2+b^2+c^2)+3}\geq2$ $\Longleftrightarrow ab+ac+bc\geq3$ which is true by AM-GM Inequality since $abc=1$ $\boxed{}$