are you absolutely sure about the letters pierre, cuz they don't seem to be symetrical

Yes, the problem is correct.

And Igor has solved it..... Pierre.

okay, i've solved it too. i just didn't want to get into computations without being sure about the statement. let's compute $PQ$. Let $I$ be the incenter, then $APIR$ and $BPIQ$ are cyclic quadrilaterals, so $\angle QPI=\angle IBQ = \angle B /2$ and $\angle IPR= \angle IAR = \angle A/2$, so $\angle QPR = \displaystyle \frac {\angle A+\angle C}2 =\frac \pi 2 - \frac { \angle B} 2$. So, using the Sines Law in the triangle $PQR$ we get $PQ = 2r \sin \displaystyle \frac \pi 2 - \frac { \angle B} 2 = 2r \cos \frac {\angle B}2$, where $r$ is the inradius of the triangle $ABC$. Thus the inequality becomes \[ \sum \frac a{ 2r\cos \frac {\angle B}2 }\geq 6 \ \Leftrightarrow \sum \frac a{\cos \frac {\angle B}2 } \geq 12r \quad (1)\] But \[\displaystyle \sum \frac a{\cos \frac {\angle B}2} \geq 3 \sqrt[3] { \frac {abc}{ \cos \frac {\angle A}2 \cos \frac {\angle B}2 \cos \frac {\angle C}2 }} \quad (2) \] so we must prove that \[{ abc \geq 4^3\cdot r^3 \cos \frac {\angle A}2 \cos \frac {\angle B}2 \cos \frac {\angle C}2 } \] \[{ 4Rrs \geq 4^3\cdot r^3 \cos \frac {\angle A}2 \cos \frac {\angle B}2 \cos \frac {\angle C}2 } \] \[ Rs \geq 16\cdot r^2 \sqrt{ \frac { s^3(s-a)(s-b)(s-c)}{(abc)^2}} \] \[ Rs \geq 16 \cdot r^2 \frac {Ss}{4RS} \ \Leftrightarrow \ R^2 \geq 4\cdot r^2 \] which is true by Euler's inequality (I used the relationships $abc=4RS = 4Rrs$, where $S,s$ are the area and semi-perimeter of the triangle $ABC$).

Let$ AB=x+y,BC=y+z,AC=x+z$ We got $PQ=\frac{2ry}{\sqrt{r^2+y^2}}$ So,$AB/PQ+BC/QR+AC/RP =(1/2ry)[(x+y)\sqrt{y^2+r^2}+.....] =(1/2)[(x+y)\sqrt{\frac{1}{y^2}+\frac{1}{r^2}}+...] =(1/2)[(x+y)\sqrt{\frac{1}{y^2}+\frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}}+...] =(1/2)[\sqrt{(x^2+xy+xy+y^2)(\frac{1}{y^2}+\frac{1}{xy}+\frac{1}{xz}+\frac{1}{xy})}+...] \geq (1/2)[(1+1+\sqrt{\frac{x}{z}}+\sqrt{\frac{x}{z}})+...] \geq 6$ here uses Cauchy and AMGM inequality and the fact $r=\sqrt\frac{xyz}{x+y+z}$ chao

The PIQB is cyclic quadrilateral so by Ptolemy theorem we have a\cdot r = PQ\cdot IB or \frac {BC}{PQ} = \frac{IB}{r} . So its enought to prove that \Sigma{IA} \geq 6r which is true by Erdos-Mordell theorem.

i forgot the paper i'll post my sol later

here is my solution: we have BC/PQ+CA/QR+AB/RP>=3(BC.CA.AB/PQ.QR.RP)^(1/3) but: PQ=BIsinB=2BQsin(B/2)where I is the incenter of ABC and similarly: QR=2CRsin(C/2); RP=2APsin(A/2). so PQ.QR.RA=BQ.CR.AP.8sin(A/2)sin(B/2)sin(C/2) but 8sin(A/2)sin(B/2)sin(C/2)<=1.(by AM-GM and jensen) and BC.CA.AB=(BQ+CR)(CR+AP)(AP+BQ)>=8BQ.CR.AP so BC.CA.AB/PQ.QR.RP>=8 and we are done.

actualy, this problem is too easy.