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Let $$f(z) = (1-1/t) z^w + z/t - 1$$ with integers $t\geq2$ and $w\geq2$.Let $r=1+1/(tw^3)$. How do I show $$\left\lvert f(r e^{i\varphi}) \right\rvert \geq \left\lvert f(r) \right\rvert$$ for any $\varphi$?

It is clear, that there is a local minimum at $\varphi=0$. The inequality is relatively easy to prove for $w=2$.

The problem is equivalent to show, that $$\cos(\varphi)-1 + (1-1/t) r^w \left(\frac{t}{r} \left(\cos(w\varphi)-1\right) - \left(\cos((w-1)\varphi)-1\right) \right) \leq 0.$$ This formulation comes from writing the square of the absolute value in terms of sin and cos.

How can I prove this?

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Given that you're asking how one could prove a certain inequality, how do you know it's true in the first place? Is it a corollary of a more general result? or are there numerical computations which suggest this inequality should be true? – Yemon Choi Aug 4 '10 at 17:56
The latter one. There are numerical computations for some (a lot of) $w$ and $t$. There the minima/maxima were calculated (numerically). I have not found a counter example yet. So I suggest, that the inequality should hold. – Daniel Krenn Aug 4 '10 at 18:12
An easy observation which may or may not be useful: If we view the set {(1−1/t) r^w e^{iwφ} + (r/t) e^{iφ} | φ∈ℝ} as a curve on the complex plane, it is an epitrochoid. Therefore the inequality in question can be viewed as a statement about which point on a given epitrochoid is the closest to a given point under certain conditions. – Tsuyoshi Ito Aug 7 '10 at 21:00

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