The title says it all, but I'm also to show that $\int_{0}^{2\pi}v(e^{it})^{4}dt \leq 36\int_{0}^{2\pi}u(e^{it})^{4}dt$, which I'm assuming will follow easily.
I have tried using the fact that $f^{4}(z) = \frac{1}{2\pi i}\int_{\gamma} \frac{f^{4}(w)}{w-z} dw$ and the Cauchy-Schwartz inequality $(\int fg)^{2} \leq (\int f^{2})(\int g^{2})$ to show some other inequalities, but none were very useful. I know that since $f^{4}$ is holomorphic on the unit sphere since $f$ is as well, its integral over $S^{1}$ will be zero. But using the Cauchy-Schwartz inequality doesn't seem to get me anywhere. Any hints? Thank you.
Note that $$0=(f(0))^4=\int_{C^+(0,1)}\frac{f(z)}{z}dz=i\int_{0}^{2\pi}( u(e^{it})+iv(e^{it}))^4dt$$ Hence $$0=\int_{0}^{2\pi}\left( u^4(e^{it})+v^4(e^{it})-6u^2(e^{it})v^2(e^{it})\right)dt$$ It follows that $$\eqalign{\int_{0}^{2\pi}u^4(e^{it})dt +\int_{0}^{2\pi}v^4(e^{it})dt&=6\int_{0}^{2\pi}u^2(e^{it})v^2(e^{it})dt\cr&\le 6 \sqrt{\int_{0}^{2\pi}u^4(e^{it})dt} \sqrt{\int_{0}^{2\pi}v^4(e^{it})dt}}$$ Now, if $$\eqalign{A&=\min\left(\int_{0}^{2\pi}u^4(e^{it})dt, \int_{0}^{2\pi}v^4(e^{it})dt\right)\cr B&=\int_{0}^{2\pi}u^4(e^{it})dt+ \int_{0}^{2\pi}v^4(e^{it})dt}$$ then we conclude from the above inequality $B\le 6\sqrt{B}\sqrt{A}$,and consequently $B\le 36 A$, that is $$\int_{0}^{2\pi}u^4(e^{it})dt+ \int_{0}^{2\pi}v^4(e^{it})dt\le 36\min\left(\int_{0}^{2\pi}u^4(e^{it})dt, \int_{0}^{2\pi}v^4(e^{it})dt\right)$$ This implies both desired inequalities.