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Show that a function analytic in the disc $\left | z \right |<1+\epsilon$ for some $\epsilon>0$ satisfies $\displaystyle f(z)=i Im(f(0))+\frac{1}{2\pi}\int_0^{2\pi}\frac{e^{it}+z}{e^{it}-z}Re(f(e^{it}))dt$

Suppose that the function above also satisfies $Re(f(e^{it}))\geq0, \forall t\in[0,2\pi]$, prove that: $\displaystyle \frac{1-\left | z \right |}{1+\left | z \right |}Re(f(0))\leq Re(f(0))\leq\frac{1+\left | z \right |}{1-\left | z \right |}Re(f(0)),\forall z: \left | z \right |<1$

I don't even know where to begin ._.

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The first line is the Schwarz integral formula, and it follows directly from the Poisson integral formula: The real part is the Poisson integral of the real part of $f$ on the circle, and the integral is analytic as a function of $z$, so the difference of $f$ and the integral is an imaginary constant. Evaluating both sides for $z=0$ you find that it has to be the imaginary part of $f(0)$.

In your line of inequalities the term in the middle should have argument $z$, not $0$. After that, the inequalities follow easily from $$ \frac{1-|z|}{1+|z|} \le \left|\frac{e^{it}+z}{e^{it}-z}\right| \le \frac{1+|z|}{1-|z|}$$ and $$u(0) = \frac1{2\pi} \int_0^{2\pi} u(e^{it}) \, dt,$$ where $u$ is the real part of $f$.

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