In my obsevation:

If $f(x)$ and $g(x)$ be two continuous funcions, and they have derivative and second derivative are also continuous in interval $[a, b]$. If $f(x)$ is more curvature than $g(x)$ in interval $[a, b]$ then length of $f(x)$ from $a$ to $b$ seem longer than length of $g(x)$ from $a$ to $b$ ?

Example: Length of any curve (curve is not a line) is longer than a line in every interval.

This observation can fomulate that:

$$\frac{|f''(x)|}{(1+f'(x)^2)^{\frac{3}{2}}} > \frac{|g''(x)|}{(1+g'(x)^2)^{\frac{3}{2}}}$$ for all $x \in [a, b]$

then $$\int_{a}^{b}{ \sqrt{1+f'(x)^2}}dx \ge \int_{a}^{b}{ \sqrt{1+g'(x)^2}}dx$$

My question: I think this is true and well-known, can you share me a reference. If this result is not known can you give a proof or a comment? Thank You very much.

In my obsevation:

If $f(x)$ and $g(x)$ be two continuous funcions, and they have derivative and second derivative are also continuous in interval $[a, b]$. If $f(x)$ is more curvature than $g(x)$ in interval $[a, b]$ and $f(a)=g(a)$, $f(b)=g(b)$ then length of $f(x)$ from $a$ to $b$ seem longer than length of $g(x)$ from $a$ to $b$ ?

Example: Length of any curve (curve is not a line) is longer than a line in every interval.

This observation can fomulate that:

$$\frac{|f''(x)|}{(1+f'(x)^2)^{\frac{3}{2}}} \ge \frac{|g''(x)|}{(1+g'(x)^2)^{\frac{3}{2}}}$$ for all $x \in [a, b]$ and $f(a)=g(a)$, $f(b)=g(b)$

then $$\int_{a}^{b}{ \sqrt{1+f'(x)^2}}dx \ge \int_{a}^{b}{ \sqrt{1+g'(x)^2}}dx$$

My question: I think this is true and well-known, can you share me a reference. If this result is not known can you give a proof or a comment? Thank You very much.