Let $z=r(\cos \varphi+i \sin \varphi)$ where $r\geq 1$. From De-Moivre's formula we have that $z^2=r^2(\cos 2\varphi+i \sin 2\varphi)$. Now we have that \begin{align*} |f(z)|&=|z^2-z+a| =|r^2(\cos {2\varphi}+i\sin 2\varphi)-r(\cos \varphi+ i\sin \varphi)+a| \\ &=|r^2 \cos 2\varphi-r\cos \varphi+a+i(r^2\sin 2\varphi-r \sin \varphi)| \\ &=\sqrt{(r^2 \cos 2\varphi-r\cos \varphi+a)^2+(r^2\sin 2\varphi-r \sin \varphi)^2} \\ &=\sqrt{r^4+r^2-2r^3(\cos 2\varphi \cdot \cos \varphi+\sin 2\varphi \cdot \sin \varphi)+a^2+2r^2 a \cos 2\varphi -2ra \cos \varphi} \\ &= \sqrt{r^4+r^2-2r^3\cos \varphi +a^2+2r^2a \cos 2 \varphi-2ra \cos \varphi } \end{align*} Now if we show that $|f(z)|^2 \geq |f(z_0)|^2$ for some $|z_0|=1$ then we'd be done because both $|f(z)|,|f(z_0)|$ are positive. Now, we will show that the function $g: [1,\infty) \to \mathbb{R}$ defined as $$g(r)=|f(z)|^2=r^4+r^2-2r^3\cos \varphi +a^2+2r^2a \cos 2 \varphi-2ra \cos \varphi $$is increasing. Note that \begin{align*} g'(r) &=4r^3+2r-6r^2\cos \varphi+4ar\cos 2 \varphi-2a\cos \varphi \\ g''(r) &=12r^2+2-12r\cos \varphi+4a\cos 2 \varphi \\ g'''(r) &=24r-12\cos \varphi \end{align*} It is clear that $g'''(r)>0$ which means that $g''(r)$ is increasing i.e $$g''(r)\geq g''(1)=14-12\cos \varphi+4a \cos 2 \varphi$$We divide in two cases if $\cos 2 \varphi \geq 0$ clearly $g''(1)>0$ if $\cos 2 \varphi <0$ and since $a<1$ then we have $g''(1)<14-12\cos \varphi+4\cos 2 \varphi$ which has a negative discriminant when substiuting $\cos 2 \varphi=2\cos ^2 \varphi-1$, wich implies $g''(1)>0$. So, we have that $g''(r)>0$ wich implies that $g'(r)$ is increasing i.e $$g'(r)\geq g'(1)=6-6\cos \varphi+4a\cos 2 \varphi-2a \cos \varphi$$. We again divide into two cases. if $2\cos 2\varphi-\cos \varphi\geq 0$, it is clear that $g'(1)>0$ since $a>0$. if $2\cos 2\varphi-\cos \varphi <0$, again substituting $\cos 2 \varphi=2\cos^2 \varphi-1$ and since $a<1$ we easily get that $g'(1)\geq (2\cos \varphi-1)^2\geq 0$. So we proved that $g(r)$ is increasing, so we have that $g(r)\geq g(1)$ wich is equivalent to $$|f(r(\cos \varphi+i\sin \varphi))|^2\geq |f(\cos \varphi+i \sin \varphi)|^2$$which proves the problem statement.