When is the moment of inertia of a smooth plane curve is maximum?

Given a smooth plane curve $(x(s),y(s))$, parameterized in arc length $s$, of fixed finite length $L$, its moment of inertia about its center of mass (axis perpendicular to the plane) is given as $$MI = \int_0^L ((x(s)-x_{cm})^2 + (y(s)-y_{cm})^2) ds$$. What I predict from earlier discussions, and almost convinced is that if we fix length $L$, $MI$ is maximum when the curve is a straight line. I lack the faculty of mathematical machinery (I guess calculus of variations) to prove it, hence is my gentle request to help me out in proving it and thoroughly understanding the situation and all the corollaries and nuances. This not just the result I need, but I want to do more with it and hence would like understand all the things that are making this result and even more general ones (only to plane curves though).

There is no need for any calculus of variation. Ordinary calculus is enough.

For simplicity of derivation, we will use complex numbers to represent points on the plane.
Let $z(s) = x(s) + i y(s)$ and WOLOG, we will assume $z(0) = 0$. We can express the position on our curve as an integral:

$$z(t) = \int_0^t z'(s)\;ds$$

Let $\theta(t) = \begin{cases} 1 & t > 0\\ 0 & t \le 0\end{cases}$ be the step function. The center of mass is given by

$$\begin{align}z_{cm} &= \frac{1}{L} \int_0^L z(t) dt = \frac{1}{L} \int_0^L \int_0^t z'(s)\;ds\; dt\\ &= \frac{1}{L} \iint_{[0,L]^2} \theta(t-s) z'(s)\;ds\; dt = \int_0^L \left(1-\frac{s}{L}\right) z'(s)\;ds \end{align} $$

The moment of inertia w.r.t. $z(0)$, the origin, is given by

$$\begin{align} \mathcal{M}_0 &= \int_0^L |z(s)|^2 ds = \int_0^L \left(\int_0^t z'(s_1)ds_1\right)\left(\int_0^t \bar{z}'(s_2) ds_2\right) dt\\ &=\iiint_{[0,L]^3} \theta(t-s_1)\theta(t-s_2) z'(s_1) \bar{z}'(s_2) ds_1 ds_2 dt\\ &=\iiint_{[0,L]^3} \theta(t-\max(s_1,s_2)) z'(s_1) \bar{z}'(s_2) ds_1 ds_2 dt\\ &= \iint_{[0,L]^2} \left(L - \max(s_1,s_2)\right) z'(s_1) \bar{z}'(s_2) ds_1 ds_2 \end{align}$$ and hence the moment of inertia w.r.t. the center of mass is

$$\mathcal{M}_{cm} = \mathcal{M}_0 - L |z_{cm}|^2 = L \iint_{[0,L]^2} \Lambda(s_1,s_2) z'(s_1) \bar{z}'(s_2) ds_1 ds_2$$

where $$\begin{align}\Lambda(s_1,s_2) &= 1 - \max\left(\frac{s_1}{L},\frac{s_2}{L}\right) - \left(1-\frac{s_1}{L}\right)\left(1-\frac{s_2}{L}\right)\\ &= \left( 1 - \max\left(\frac{s_1}{L},\frac{s_2}{L}\right)\right)\min\left(\frac{s_1}{L},\frac{s_2}{L}\right) \end{align}$$

Since $|z'(s)| \equiv 1$ and $\Lambda(s_1,s_2) > 0$ for $(s_1,s_2) \in (0,L)^2$, we can bound $\mathcal{M}_{cm}$ as

$$\mathcal{M}_{cm} \le L \iint_{[0,L]^2} \Lambda(s_1,s_2) |z'(s_1)||z'(s_2)| ds_1 ds_2 = L \iint_{[0,L]^2} \Lambda(s_1,s_2) ds_1 ds_2$$ Notice the equality in above inequality is achieved when and only when $z'(s)$ is a constant.
We can conclude $\mathcal{M}_{cm}$ is largest for straight lines.

Assumptions:
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1. the center of mass is $0$ $$ 0=\int_0^L\delta f\,\mathrm{d}s\tag{1} $$ 2. $f$ is parametrized by arc length: $f'\cdot f'=1$ $$ \begin{align} 0&=\int_0^Lf'\cdot\delta f'\,\mathrm{d}s\\ &=\int_0^Lf'\cdot\,\mathrm{d}\delta f\\ &=\Big[\,f'\cdot\delta f\,\Big]_0^L-\int_0^Lf''\cdot\delta f\,\mathrm{d}s\\ &=-\int_0^Lf''\cdot\delta f\,\mathrm{d}s\tag{2} \end{align} $$ Note that $\delta f$ can be adjusted in the direction of $f'$ near $0$ and $L$ without affecting the integral of $f''\cdot\delta f$ since $f'\cdot f''=0$. Thus, we can assume $\Big[\,f'\cdot\delta f\,\Big]_0^L=0$.
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3. the moment of inertia is stationary $$ 0=\int_0^Lf\cdot\delta f\,\mathrm{d}s\tag{3} $$ Conclusions:

For an extreme $f$, any $\delta f$ that satisfies $(1)$ and $(2)$ also satisfies $(3)$; thus, linearity says that there are constants $a$ and $b$ so that $$ f=a+bf''\tag{4} $$ Since $f$ is parameterized by arclength $f'\cdot f''=0$, integrating the dot product of $(4)$ with $f'$ yields $$ \frac12f\cdot f=a\cdot f+c\tag{5} $$ Equation $(5)$ represents an arc of the circle $$ |f-a|=\left(2c+|a|^2\right)^{1/2}\tag{6} $$ or in the extreme case, a line segment.

Checking Possible Arcs:

The equation of an arc of radius $r$ is $$ f=r(\cos(s/r),\sin(s/r))\tag{7} $$ The center of mass is $$ \frac1L\int_{-L/2}^{L/2}r(\cos(s/r),\sin(s/r))\,\mathrm{d}s=\left(\frac{2r^2}{L}\sin\left(\frac{L}{2r}\right),0\right)\tag{8} $$ The moment of inertia is $$ \begin{align} &\frac{r^2}{L}\int_{-L/2}^{L/2}\left[\left(\cos(s/r)-\frac{2r}{L}\sin\left(\frac{L}{2r}\right)\right)^2+\sin^2(s/r)\right]\,\mathrm{d}s\\ &=\frac{r^2}{L}\int_{-L/2}^{L/2}\left[1-\frac{4r}{L}\sin\left(\frac{L}{2r}\right)\cos(s/r)+\frac{4r^2}{L^2}\sin^2\left(\frac{L}{2r}\right)\right]\,\mathrm{d}s\\ &=r^2-\frac{4r^4}{L^2}\sin^2\left(\frac{L}{2r}\right)\\ &=r^2\left(1-\frac{\sin^2\left(\frac{L}{2r}\right)}{\left(\frac{L}{2r}\right)^2}\right)\tag{9} \end{align} $$ $(9)$ increases to $\frac{L^2}{12}$ as $r\to\infty$. Thus, the maximal moment of inertia would be at the extreme case of a line segment.