We know that if triangle $ABC$ has length of sides are $a,b,c$ then its area equal to $\frac{2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}-a^{4}-b^{4}-c^{4}}{4}$. And if $a,b,c$ are positive reals such that $2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}-a^{4}-b^{4}-c^{4}>0$ then $a,b,c$ are lengths of sides of a triangle. Now, WLOG we can assume that $a=1$. We need prove that : Exist positive reals $b,c$ such that $0<2b^{2}+2b^{2}c^{2}+2c^{2}-1-b^{4}-c^{4}=2p^{2}q^{2}b^{2}+2q^{2}r^{2}b^{2}c^{2}+2p^{2}r^{2}c^{2}-p^{4}-q^{4}b^{4}-r^{4}c^{4}.$ At this time, that is all what i know.

Are you kidding me? The problem does not require $p, q $ and $ r$ to be distinct. So let $p = q = r = 1$ and $1$ lies in the interval $ (\frac{2}{5},\frac{5}{2}) $. The two triangles are congruent because $pa = a, qb = b, rc = c$ and have the same area. Done and totally acceptable! On the other hand, if $p, q $ and $ r$ are distinct, arbitrarily let $r$ = 1 and $ \frac{1}{p} = q$, the two triangles to be $ABC$ and $A'B'C'$ with $AB$ = $a$, $BC$ = $c$, $CA$ = $b$, and $A'B'$ = $pa$, $B'C'$ = $rc$ = $c$, $C'A'$ = $qb$. We can also let $ \angle BAC$ = $ \angle B'A'C'$ and $Sx$ and $Sy$ be the areas of $ABC$ and $A'B'C'$, respectively. We have $Sx$ = $ \frac {1}{2}ab sin\angle BAC$ = $ \frac {1}{2}paqb sin\angle B'A'C'$ = $Sy$. Actually, the interval range in the problem is not really important; to be nice you can use two numbers that are inverses of each other.