Let's reduce $x$ in terms of triangle $ABC$. The diagram reminds us of Pompeiu's theorem, so let's use that. Rotate the diagram on the right by $60^\circ$ counterclockwise around $R$. Let $M,Q$ map to $M',Q'$ respectively. Also note that $P,R$ map to $Q,R$ respectively. $\angle MRM'=60^\circ$ because that's how much we rotated it by. It's also isosceles with $MR=RM'$, so $MM'R$ is equilateral. Then $MM'=c$. By rotation, $M'Q=a$, and $MQ$ is still equal to $b$, so by SSS $ABC$ is congruent to $MM'Q$. So to construct $x$, we can just draw an equilateral triangle $RAB$ with $R$ and $C$ on opposite sides of $AB$. Then $RC=x$. But then sides $AC$ and $BC$ would be unnecessary. Ooh it looks like $R,D,C$ are collinear. Since $\angle ADC=120^\circ$, we have to show that $\angle RDA=60^\circ$. Let's construct stuff from $RAB$. $\angle ADB=120^\circ$, so it's on some circle with arc $AB$. Some experimentation motivates $R,A,D,B$ being a cyclic quadrilateral. It's true because $\angle RAB+\angle ADB=60+120=180^\circ$. But then $\angle RDA=\angle RBA=60^\circ$, so $R,D,C$ are collinear. Let's think about what we want to prove again. We want to show that $RC=x=u+v+w=DA+DB+DC$, but $R,D,C$ are collinear, so this is equivalent to $RD=DA+DB$. When you got nothing just use Pompeiu's theorem construction again. Rotate $RAB$ counterclockwise by $60^\circ$. Let $B,D$ go to $B',D'$ respectively. Ooh $D,B,D'$ look collinear. By rotation $\angle ABD+\angle RBA+\angle RBB'+\angle B'BD'=\angle ABD+60+60+\angle BAD=60+60+(180-\angle ADB)=180^\circ$, so yes. Focus on the results of Pompeiu. By rotation, $RD=RD'$, and $\angle DRD'=60^\circ$, so $RDD'$ is equilateral. Then $RD=DD'=DB+BD'=DB+DA$, so we're done.

Construct equilateral triangle $XAB$ such that $D$ and $C$ are on opposite sides of $AB$. Rotate the diagram $60^\circ$ counterclockwise such that $X,A,B,C,D$ go to $X,B,B',C',D'$, respectively. By rotation, $\angle DBD'=\angle DBA+\angle ABX+\angle XBB'+\angle B'BD'=\angle DBA+60+60+\angle BAD=120+(180-\angle ADB)=180$, so $D,B,D'$ are collinear. Also, $XD=XD'$, and $\angle DXD'=60^\circ$, so $XDD'$ is equilateral. This means $XD=DD'=DB+BD'=DB+DA=u+v$. Furthermore $\angle AXB+\angle ADB=60+120=180^\circ$, so $XADB$ is a cyclic quadrilateral. Then $\angle ADC+\angle ADX=120+\angle ABX=120+60=180^\circ$, so $X,D,C$ are collinear, which means $XC=XD+DC=u+v+w$. Rotation tells us that $\angle CXC'=60^\circ$, and $XC=XC'$, so $XCC'$ is equilateral. However, $BX=AB=c$, $BC=b$, and $BC'=AC=a$, so $CXC'$ is congruent to $PQR$. Then $x=PQ=XC=u+v+w$, which completes the solution.