This is a neat trick using Cauchy's theorem for the function $\frac{f(z)^{2k}}z$, i.e., $ 0 = \int_{|z|=r} \frac{f(z)^{2k}}z \, dz = \int_0^{2\pi} (u(re^{i\theta})+iv(re^{i\theta}))^{2k} \, d\theta. $ Then multiply it out, take the real part, throw the integral of $v^{2k}$ on one side and get rid of all the negative terms to get the inequality (where for ease of notation $u=u(re^i\theta)$ and $v=v(re^{i\theta})$) $ \int_0^{2\pi} v^{2k} \, d\theta < \int_0^{2\pi}\left( { 2k \choose 2} u^{2}v^{2k-2} + {2k \choose 6} u^{6} v^{2k-6} + \ldots \right) \, d\theta. $ Now for $2 \le j \le 2k-2$, Hölder's inequality with $p=\frac{2k}{j}$ and $q=\frac{2k}{2k-j}$ gives $ \int_0^{2\pi} u^{j} v^{2k-j} \, d\theta \le \left( \int_0^{2\pi} u^{2k} \, d\theta \right)^{\frac{j}{2k}} \left( \int_0^{2\pi} v^{2k} \, d\theta \right)^{\frac{2k-j}{2k}} $ Note that this inequality is trivially true if $j=2k$. Writing $ U = \left( \int_0^{2\pi} u^{2k} \, d\theta \right)^{\frac1{2k}} \quad \text{ and } \quad V = \left( \int_0^{2\pi} v^{2k} \, d\theta \right)^{\frac1{2k}} $ we get $ V^{2k} < { 2k \choose 2 } U^{2} V^{2k-2} + { 2k \choose 6 } U^{6} V^{2k-6} + \ldots $ and dividing both sides by $U^{2k}$ and writing $X=V/U$ we have $ X^{2k} < { 2k \choose 2 } X^{2k-2} + { 2k \choose 6 } X^{2k-6} + \ldots $ Now the left-hand side is a polynomial of degree $2k$ in $X$, whereas the right-hand side is a polynomial of degree $2k-2$ in $X$, so if $M_k$ denotes the largest solution of the corresponding polynomial equality, then we know that $X < M_k$. Note that $M_k$ depends only on the coefficients of the polynomial, not on $f$ or $r$. To get the claimed inequality, set $c_k = M_k^{2k}$.
I shamelessly stole this proof from Marcel Riesz's original paper Marcel Riesz, Sur les fonctions conjuguées, Math. Z. 27 (1928), no. 1, 218–244 (The proof appears on page 221.)