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Notice that $g(t), 0\le t\le 2\pi$, is a closed curve in the plane. It is parametrized with respect to arc length, and its curvature is given by $|f'(t)|$. By assumption, $a\le |f'|\le b$.

For convenience, let's discuss the case $x=0$. I claim that the inequality may be viewed as the condition that we will still be able to return $g(0)$, given the restrictions on the curvature.

This is not really a complete solution, it contains possibly challenging exercises to the reader, but I have some hope that one could make a full proof out of this. I'll also only address the lower bound.

The function $g(t), 0\le t\le 2\pi$, defines a closed curve in the plane. It is parametrized with respect to arc length, and its curvature is given by $f'(t)$. By assumption, $a\le f'\le b$ (so the curve is also convex).

For the lower bound, we are interested in how small $|g(x)-g(0)|$ can become under these restrictions: curvature $\ge a$, the piece in question has length $x$, and the total length equals $2\pi$.

It seems intuitively plausible that the minimum is achieved for an American football type curve that we obtain by gluing together two circular arcs of radius $R=1/a$ and length $\pi$ each. (I don't know how to prove it formally though.) This curve is not smooth at the gluing points, but of course we can approximate by smooth curves, so this is not an issue.

The minimal distance $|g(x)-g(0)|$ is achieved by considering distances parallel to the small axis of symmetry of our football. Notice also that if we brought these two points $g(0),g(x)$ closer together still by moving (parts of) our circular arcs around, then it would not be possible to close the curve with another arc of length $2\pi -x$ and at the same time keep the curvature $\ge a$ throughout. (We of course can't prove the claim without making use of this non-local feature of the problem.)

If all this is swallowed, then the rest follows by direct computation. By elementary geometry, $$ G(x):=|g(x)-g(0)| = 2R \left( \cos\frac{\pi -x}{2R} - \cos \frac{\pi}{2R} \right) . $$ We want to show that this is $\ge$ $$ H(x):=a|e^{ix}-1| = \frac{2}{R}\, \sin (x/2) . $$ For $x=\pi$, this boils down to showing that $$ 2R^2 \sin^2 \frac{\pi}{4R} \ge 1 , $$ or, equivalently, $\sin\delta\ge (\sqrt{8}/\pi)\delta$ for $0<\delta\le \pi/4$, which (just about) works. The upper bound on $\delta$ came from the fact that $R=1/a\ge 1$.

I have shown that $H(\pi)\ge G(\pi)$, and to obtain that $H(x)\ge G(x)$ also for general $0<x<\pi$, I can compare the derivatives: it's not hard to check that $H'\le G'$ in this range for arbitrary $R\ge 1$.

Notice that $g(t), 0\le t\le 2\pi$, is a closed curve in the plane. It is parametrized with respect to arc length, and its curvature is given by $|f'(t)|$. By assumption, $a\le |f'|\le b$.

For convenience, let's discuss the case $x=0$. I claim that the inequality may be viewed as the condition that we will still be able to return $g(0)$, given the restrictions on the curvature.

The comparison curve $e^{it}$ is the unit circle, and if we consider the points corresponding to $t=0$, $t=x$ and the center of this circle, we obtain an isosceles triangle with base of length $|e^{ix}-1|$ and legs of length $1$.

Now compare this with the corresponding points $g(x)$, $g(0)=0$. The maximum curvature we are allowed to use is $b$, so we have to be able to return from $g(x)$ to $g(0)$ along an arc of length $2\pi -x$. To minimize the length needed, we should make the curvature large, at least initially, but we are not allowed to go beyond $b$.

This suggests a corresponding isosceles triangle with base $|g(x)-g(0)|$ and legs of length $1/b$. If your (second) inequality holds with equality, this triangle is similar to the one from above, and we are just about able to return to $g(0)$. If $|g(x)-g(0)|$ became larger still, we would run into a contradiction.

The first inequality can be proved similarly, of course.

Notice that $g(t), 0\le t\le 2\pi$, is a closed curve in the plane. It is parametrized with respect to arc length, and its curvature is given by $|f'(t)|$. By assumption, $a\le |f'|\le b$.

For convenience, let's discuss the case $x=0$. I claim that the inequality may be viewed as the condition that we will still be able to return $g(0)$, given the restrictions on the curvature.

Notice that $g(t), 0\le t\le 2\pi$, is a closed curve in the plane. It is parametrized with respect to arc length, and its curvature is given by $|f'(t)|$. By assumption, $a\le |f'|\le b$.

For convenience, let's discuss the case $x=0$. I claim that the inequality may be viewed as the condition that we will still be able to return $g(0)$, given the restrictions on the curvature.

The comparison curve $e^{it}$ is the unit circle, and if we consider the points corresponding to $t=0$, $t=x$ and the center of this circle, we obtain an isosceles triangle with base of length $|e^{ix}-1|$ and legs of length $1$.

Now compare this with the corresponding points $g(x)$, $g(0)=0$. The maximum curvature we are allowed to use is $b$, so we have to be able to return from $g(x)$ to $g(0)$ along a circular arc of radius $\ge 1/b$ and of length at most $2\pi -x$ (recall that $g(t)$ has total length $2\pi$, and length $x$ has been used up to go from $g(0)$ to $g(x)$). This suggests a corresponding isosceles triangle with base $|g(x)-g(0)|$ and legs of length $1/b$. If your (second) inequality holds with equality, this triangle is similar to the one from above, and we are just about able to return to $g(0)$. If $|g(x)-g(0)|$ became larger still, we would run into a contradiction.

The first inequality can be proved similarly, of course.

Notice that $g(t), 0\le t\le 2\pi$, is a closed curve in the plane. It is parametrized with respect to arc length, and its curvature is given by $|f'(t)|$. By assumption, $a\le |f'|\le b$.

For convenience, let's discuss the case $x=0$. I claim that the inequality may be viewed as the condition that we will still be able to return $g(0)$, given the restrictions on the curvature.

The comparison curve $e^{it}$ is the unit circle, and if we consider the points corresponding to $t=0$, $t=x$ and the center of this circle, we obtain an isosceles triangle with base of length $|e^{ix}-1|$ and legs of length $1$.

Now compare this with the corresponding points $g(x)$, $g(0)=0$. The maximum curvature we are allowed to use is $b$, so we have to be able to return from $g(x)$ to $g(0)$ along an arc of length $2\pi -x$. To minimize the length needed, we should make the curvature large, at least initially, but we are not allowed to go beyond $b$. 

This suggests a corresponding isosceles triangle with base $|g(x)-g(0)|$ and legs of length $1/b$. If your (second) inequality holds with equality, this triangle is similar to the one from above, and we are just about able to return to $g(0)$. If $|g(x)-g(0)|$ became larger still, we would run into a contradiction.

The first inequality can be proved similarly, of course.