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Clearly if $p=q$ or $a=b$ you can conclude to the disired result. Only remains to examine the case when $a \not= b \not= c \not= a$ and $p \not= q \not= r \not= p$

Any ideas?

This problem is only tricky. Suppose $p=q=r$ is false. WLOG, let us assume that $a\geq b,a\geq c$. Call a triple $(x,y,z) $ increasing iff $x\leq y\leq z$ and decreasing iff $x\geq y\geq z$. I claim that the set {$(p,q,r),(q,r,p),(r,p,q) $} contains an increasing or a decreasing triple. Suppose it does not contain an increasing triple then {$(r,q,p),(p,r,q),(q,p,r) $} contains an increasing triple, assume $(r,q,p) $ is increasing then $(p,q,r) $ is decreasing hence our required set does have a decreasing triple if it does not have an increasing triple. Now Rearrangement Inequality does the job.