Answer
The are length is not less than $5 / 3$
Work Step by Step
Since $f(x)=x^{4 / 3},$ we have
$$
f^{\prime}(x)=\frac{4}{3} x^{1 / 3}
$$
for $x \geq 1,$ we have $x^{1 / 3} \geq 1$; therefore
$$
\left.1+f^{\prime}(x)\right)^{2} \geq 1+\left(\frac{4}{3}\right)^{2}=\left(\frac{5}{3}\right)^{2}
$$
By comparison, we get
\begin{align*}\int_{1}^{2} \sqrt{\left.1+f^{\prime}(x)\right)^{2}} d x &\geq \int_{1}^{2} \sqrt{\left(\frac{5}{3}\right)^{2}} d x\\
&=\frac{5}{3}[x]_{1}^{2}\\
&=\frac{5}{3}\end{align*}
Therefore, the arc length is not less than $5 / 3$
