Using Mean Value Theorem to solve an inequality involving $\cos^{-1}(x)$ Question:
Using the Mean Value Theorem, show that for all $-\frac{1}{2}\lt a,b \lt \frac{1}{2}$ with $a\lt b$
$$\lvert \cos^{-1}(a)-\cos^{-1}(b)\rvert \lt \frac{2\sqrt{3}}{3}\lvert a-b\rvert$$

My attempt:
Let $f(x)= \cos^{-1}(x)$
$\cos^{-1}(x)\ \text{is continuous and differentiable on the interval so by the MVT,} $
$$\lvert \cos^{-1}(b)-\cos^{-1}(a) \rvert= \lvert b-a\rvert.\lvert f'(c)\rvert$$
$$\lvert f'(c) \rvert= \frac{1}{\sqrt{1-c^2}} \lt \frac{2\sqrt{3}}{3} $$
Therefore we have,
$$\lvert \cos^{-1}(a)-\cos^{-1}(b) \rvert= \lvert a-b\rvert.\lvert f'(c) \rvert \le 1.\lvert a-b\rvert \lt \lvert b-a\rvert \frac{2\sqrt{3}}{3}$$
Because $\lvert a-b\rvert= \lvert b -a\rvert$
Would this be correct?
No, this wouldn't be correct. The equality you wrote from the MVT is not right; if you take absolute values, then you must also have $|f'(c)|$ in the RHS; and just because $f'(c) &lt; 1$ it doesn't mean its absolute value is also $&lt;1$.
$|f'(c)| = \dfrac{1}{\sqrt{1 - c^2}} &lt; \dfrac{1}{\sqrt{1 - (1/2)^2}}$ since $c^2 &lt; (1/2)^2$.