Let $P(x)=f(x)-1,Q(x)=f(x)-2$.We have that $P(x)$ has 4 real roots $x_1,x_2,x_3,x_4$ satisfy $x_1+x_2=x_3+x_4$. Let $x_1+x_2=x_3+x_4=p$. By Vieta's relations we have $x_1+x_2+x_3+x_4=-a\Leftrightarrow 2p=-a$ and $\sum_{1\leq i<j\leq 4} x_ix_j=b\Leftrightarrow x_1x_2+x_3+x_4+p^2=b\Leftrightarrow x_1x_2+x_3x_4=b-p^2=b-\dfrac{a^2}{4}$ and $\sum_{1\leq i< j< k\leq 4}x_ix_jx_k=-c\Leftrightarrow x_2x_1(x_3+x_4)+x_3x_4(x_1+x_2)=-c\Leftrightarrow p(x_1x_2+x_3x_4)=-c\Leftrightarrow$ $\Leftrightarrow p(b-\dfrac{a^2}{4})=-c\Leftrightarrow \dfrac{a}{2}(b-\dfrac{a^2}{4})=c\,\,\,(*)$ Let $r_1,r_2,r_3,r_4$ the roots of $Q(x)$.Let $r_1+r_2=m,r_3+r_4=n$ By Vieta's relations we have $m+n=-a$ and $r_1r_3+r_3r_4+mn=b$ and $r_1r_2n+r_3r_4m=-c$. By $(*)$ we have that $(r_1r_2+r_3r_4+mn-\dfrac{(m+n)^2}{4})\dfrac{m+n}{2}=(r_1r_2n+r_3r_4m)\Leftrightarrow $ $\Leftrightarrow mr_1r_2+mr_3r_4+nr_1r_2+nr_3r_4+(m+n)\dfrac{4mn-(m+n)^2}{4}=2r_1r_2n+2r_3r_4m\Leftrightarrow$ $\Leftrightarrow -(m+n)\dfrac{(m-n)^2}{4}=(n-m)(r_1r_2-r_3r_4)$. If $m-n=0$ we are done.Otherwise we have $(m+n)(m-n)=4r_1r_2-4r_3r_4\Leftrightarrow$ $\Leftrightarrow (r_1+r_2)^2-(r_3+r_4)^2=4r_1r_2-4r_3r_4 \Leftrightarrow (r_1-r_2)^2=(r_3-r_4)^2\Leftrightarrow |r_1-r_2|=|r_4-r_3|$ which implies that $r_1+r_3=r_4+r_2$ or $r_1+r_4=r_2+r_3$,QED