How does the triangle inequality work here?

Question

I was reading a proof about a complex contour integral recently, and saw a step which was a bit like $$\left|\frac1{R^2 e^{2i\theta}+1}\right|\le\frac1{R^2-1}$$ I tried to derive this. $$\left|\frac1{R^2e^{2i\theta}+1}\right|=\frac{\left|R^2e^{-2i\theta}+1\right|}{R^4+2R^2\cos2\theta+1}\le\frac{R^2+1}{R^4+2R^2\cos2\theta+1}$$But then I got stuck because the denominator does not factorise to cancel the numerator. How can this be shown?

Thanks.

Answer 1

One form of the triangle inequality is that $|x + y| \ge |x| - |y|$; prove this by applying the usual inequality to $|x| = |x + y - y|$. Then we have

$$|R^2 e^{2i\theta} + 1| \ge |R^2 e^{2i\theta}| - |1| = R^2 - 1.$$

Invert to get the desired result.

Answer 2

$|R^{2}e^{2i\theta}+1|\geq |R^{2}e^{2i\theta}|-1=R^{2}-1$ and then flipping over.

Answer 3

The locus of points in the denominator is a circle of radius $R^2$ centered at $1+0i$. What's the closest such a point can be to the origin?

Note: this geometric argument tells you the inequality is wrong if $R^2 < 1$.