logarithmic inequality

[man111]

if $a\;,b\;c\geq 2\;,$ then Minimum value of $P=\log_{a+b}c+\log_{b+c}a+\log_{c+a}b$

[ins-]

http://imomath.com/othercomp/Bul/BulMO366.pdf see problem 2.
You can use Nesbit inequality.

[man111]

but ins it is unsolved

[ins-]

Yes. It is not solved (even in the books dedicated to bulgarian math olympiad) but it isn't difficult to solve it.
It is not difficult do see that $a+b \ge 2\sqrt{ab}$.
Using this and some logarithms' properties http://www.andrews.edu/~calkins/math/we ... numb17.htm
like reciprocal property and base property 3 from "Four base properties" you can get an inequality with three variables $log_{a}b=u, log_{b}c=v, log_{c}a=w$ it is a Nesbit inequality for u,v,w.

[mihiv]

Действительно всё сводится к неравенству Nesbitt'а,которое для положительных $x,y,z$ имеет вид:$\sum \limits _{cycl (x,y,z)}\dfrac x{y+z}\geq \dfrac32$.
Покажем,сначала,что при $x,y\geq 2,\ln (x+y)\leq \ln x+\ln y.$
Предположим $x\geq y$,тогда:

$$\ln (x+y)\leq \ln (2x)=\ln 2+\ln y\leq \ln y+\ln x \qquad (1)$$

С помощью (1) получим:

$$\sum \limits _{cycl(a,b,c)}\log _{a+b}c=\sum \limits _{cycl(a,b,c)}\dfrac {\ln c}{\ln (a+b)}\geq \sum \limits _{cycl(a,b,c)}\dfrac {\ln c}{\ln a+\ln b}\geq \dfrac 32\qquad (2)$$

Заметим теперь,что первая сумма в (2) равна $\dfrac 32$ при $a=b=c=2$.

[ins-]

I think Bulgaria as well as Russia is a country of great mathematicians. My favorite problems of this kind are Canadian MO 1995 inequality as well as Bulgarian MO 2001 - Regional Round equation and some Russian problems.

[man111]

thanks mihiv and ins