I claim the answer is $c=\pm 1/2$. This is easily seen to work. Next, note that $f(c)f(-c)\ge f(a)$ iff $g(b)\ge 0$ where \[ g(b) = b^2 + (2c^2-1)b +c^4 -4a^2c^2 + a^2. \]We want the quadratic $g(b)\ge 0$ for all choices of $a,b$. This holds iff its discriminant is non-positive: $(2c^2-1)^2 - 4(c^4+a^2-4a^2c^2) = (4a^2-1)(4c^2-1)\le 0$. It is now clear that unless $4c^2=1$, one can choose $a$ such that ${\rm sgn}(4a^2-1)={\rm sgn}(4c^2-1)$ and make the discriminant positive, which would yield a contradiction. Thus, $c=\pm 1/2$ as claimed.