# Inequality concerning absolute value of a polynomial

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?

• 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

Proof: It suffices to prove that $$f(x) = \cos x + a \cos wx - b (\cos wx \cos x + \sin wx \sin x) - (1+a-b)\le 0$$ where $$a = (t-1) r^{w-1}, \ b = (1-\frac{1}{t})r^w$$. We will use the following facts: $$0 < b < 1 < a, \quad \cos \frac{\pi}{w} < 1+2a - 2b - 2ab < 1. \tag{1}$$ We put the proof at the end.

Clearly, we may restrict to $$x\in [0, \pi]$$. We split into three cases:

1. $$x \in [\arccos(1+2a-2b-2ab),\, \pi]$$: We have $$f(x) \le \cos x + \sqrt{(a-b\cos x)^2 + (b\sin x)^2} - (1+a-b)\le 0$$ $$\Longleftarrow (a-b\cos x)^2 + (b\sin x)^2 \le (1+a-b-\cos x)^2$$ $$\Longleftarrow (1-\cos x)(1+2a-2b-2ab-\cos x)\ge 0$$ $$\Longleftarrow \cos x \le 1+2a-2b-2ab$$ $$\Longleftarrow x \in [\arccos(1+2a-2b-2ab), \pi].$$

2. $$x \in [\frac{1}{w}\arccos\frac{a^2 + 2a-2b-2ab}{a^2},\, \arccos(1+2a-2b-2ab))$$: From (1), we have $$w\arccos(1+2a-2b-2ab) < \pi$$. We have $$f(x) \le a\cos wx + \sqrt{(1-b\cos wx)^2 + (b\sin wx)^2} - (1+a-b)\le 0$$ $$\Longleftarrow (1-b\cos wx)^2 + (b\sin wx)^2 \le (1+a-b-a\cos wx)^2$$ $$\Longleftarrow (1-\cos wx)(a^2 + 2a - 2b - 2ab - a^2\cos wx)\ge 0$$ $$\Longleftarrow \cos wx \le \frac{a^2 + 2a-2b-2ab}{a^2}$$ $$\Longleftarrow x \in \Big[\frac{1}{w}\arccos\frac{a^2 + 2a-2b-2ab}{a^2},\, \arccos(1+2a-2b-2ab)\Big).$$

3. $$x\in [0,\, \frac{1}{w}\arccos\frac{a^2 + 2a-2b-2ab}{a^2})$$: From (1), since $$a > b$$ and $$0\le (w-1)x \le wx \le \frac{\pi}{2}$$, we have $$f'(x) = -\sin x - aw\sin wx + b(w-1)\sin (w-1)x \le 0.$$ Since $$f(0) = 0$$, we have $$f(x) \le 0.$$ This completes the proof.

$$\phantom{2}$$

Proof of (1): We only prove $$\cos \frac{\pi}{w} < 1 + 2a-2b - 2ab$$ because the others are simple. We will use the following facts: \begin{align} 1-\frac{2}{w^2}&\ge \cos\frac{\pi}{w}, \qquad\qquad\qquad\qquad\qquad\quad (2)\\ \quad 1 + \frac{1}{tw^2} + \frac{1}{2t^2w^4} &\ge \Big(1 + \frac{1}{tw^3}\Big)^w. \qquad\qquad\qquad\qquad\ (3) \end{align} The proof is not hard and thus omitted.

According to (2), it suffices to prove that $$0 \le \frac{1}{w^2} + a-b - ab$$ or $$\frac{1}{w^2} - \frac{t-1}{t} r^{w-1}(r + (t-1)r^w -t) \ge 0.$$ According to (3), noting also that $$\frac{t-1}{t} < 1$$ and $$r + (t-1)r^w -t \ge 1 + (t-1) - t = 0$$, it suffices to prove that $$\frac{1}{w^2} - \frac{1}{r}\Big(1 + \frac{1}{tw^2} + \frac{1}{2t^2w^4}\Big) \Big[r + (t-1)\Big(1 + \frac{1}{tw^2} + \frac{1}{2t^2w^4}\Big) -t\Big]\ge 0$$ or $$\frac{(4 w^6-4 w^5-6 w^4+4 w^3) t^3+(6 w^4-4 w^3-4 w^2) t^2+(4 w^2-2 w-1) t + 1}{4t^3w^5(tw^3+1)}\ge 0$$ which is true since $$4 w^6-4 w^5-6 w^4+4 w^3\ge 0, \ 6 w^4-4 w^3-4 w^2 \ge 0, \ 4 w^2-2 w-1 \ge 0.$$ This completes the proof of (1).