Triangle inequality in ML inequality

Question

Show that $\left|\int_C \frac{1}{z^2+1}\le\frac{1}{2\sqrt{5}}\right|$, where $C$ is the straight line segment from $z=2$ to $z=2+i$.

The contour length here is 1, so all we have to worry about is the M's value to get ML. We have to maximise the thing so we have to minimise the denominator using triangle inequality.

but when I do it, the answer is not matching with what I have to prove.

|z^2+1|≥||z^2|-1|

and since we need the minimum value we use z=2.

so 4-1 = 3.

I am getting 1/3, but the answer I have to prove is different.

Answer 1

Hint 1 For two complex numbers $z$ and $w$ (with $|z+w|>0$), we have $$\dfrac{1}{|z+w|}\le\dfrac{1}{||z|-|w||}.\tag{1}$$

Hint 2 Split $C$ into two straight line segments: $C_1$ from $2$ to the midpoint $m=2+i/2$ and $C_2$ from $m$ to $2+i$.

Added. For $z_1=x_1+iy_1\in C_1$, we have $|z_1|>2 $ and for $z_2=x_2+iy_2\in C_2$, we have $|z_2|>\frac{\sqrt{17}}{2} $. Now use $(1)$, for $w=1,z=z_1^2$:

$$\frac{1}{|z_1^2+1|} \le \frac{1}{||z_1^2|-1 |}=\frac{1}{||z_1|^2-1| |}\le \frac{1}{3},\tag{2}$$

and similarly to $$\frac{1}{|z_2^2+1|} \le\frac{4}{13}.\tag{3}$$ Finally, apply ML inequality to each integral (with $L=1/2$): $$\left|\int_{C_1} \frac{1}{z_1^2+1}\right|\le\frac{1}{6} \qquad\text{ and }\qquad\left|\int_{C_2} \frac{1}{z_2^2+1}\right|\le\frac{2}{13}.\tag{4}$$

Hence $$\left|\int_{C} \frac{1}{z^2+1}\right|\le\frac{1}{6}+\frac{2}{3}\le\frac{1}{2\sqrt{5}}.\tag{5}$$

Answer 2

The length is (obviously) $1$ by the Distance Formula: $$ L = \sqrt{(2-2)^2 + (1-0)^2} = 1. $$

On $C$, parameterize $z(t) = 2+t(2+i-2) = 2+it$ where $t \in [0,1]$.

A nice way to minimize $\left|z^2+1\right|$ is by

$$ \left|z^{2}+1\right|=\left|z-i\right|\left|z+i\right|=\left|2+it-i\right|\left|2+it+i\right|=\sqrt{t^{2}-2t+5}\sqrt{t^{2}+2t+5}. $$

We can employ the First Derivative Test for both expressions. For the first one,

$$ \frac{d}{dt}\sqrt{t^{2}-2t+5}=\frac{t-1}{\sqrt{t^{2}-2t+5}}. $$

The only critical number we really need is $t=1$. Then we plug in the infimum and supremum of the domain of $t$ like

$$ \sqrt{0^{2}-2\left(0\right)+5}=\sqrt{5} \text{ and } \sqrt{1^{2}-2\left(1\right)+5}=2. $$

Hence, $\min\left\{\sqrt{t^2-2t+5}\right\} = 2$.

Do the same for the other expression to get $\min\left\{\sqrt{t^2+2t+5}\right\} = \sqrt{5}$.

Thus,

$$ |f(z)| = \left|\frac{1}{z^2+1}\right| \leq \frac{1}{2\sqrt{5}}. $$

We conclude with

$$ \left|\int_C f(z)dz\right| \leq \frac{1}{2\sqrt{5}} \cdot 1 = \frac{1}{2\sqrt{5}}. $$