(x+ny)/ny=(y+nz)/nz=(z+nx)/nx= proportiiderivate le adun numitorii intre ei si numaratorii intre ei si apoi dau factor comun ***=((x+y+z)(n+1))/((x+y+z)*n)= simplific prin (x+y+z)=(n+1)/n
atunci
x+ny=ny*(n+1)/n=y*(n+1)⇒x+ny=ny+y⇒x=y (1)
y+nz=nz*(n+1)/n=z(n+1)⇒y+nz=nz+z⇒y=z (2)
din (1) si (2)⇒x=y=z C.C.T.D.
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adunand numaratorii intre ei obtinem
x+ny+y+nz+z+nx=x+y+z+nx+ny+nz= x+y+z+ n(x+y+z)=(x+y+z)(n+1)
si, adunand numitorii intre ei, obtinem
nx+ny+nz=n(x+y+z)
am folosit proprietatea proportiilor
a/x=b/y=c/z= (a+b+c)/(x+y+z)
