A straight line through the point (h, k) where h > 0 and k > 0, makes positive intercepts on the coordinate axes. Then the minimum length of the line intercepted between the coordinate axes, is

$\left(h^{2 / 3}+k^{2 / 3}\right)^{3 / 2}$

Let the straight line be

$y-k=m(x-h)$, where m < 0

It meets X and Y axes at $A(h-k / m, 0)$ and $B(0, k-m h)$

∴  $A B^2=\left(h-\frac{k}{m}\right)^2+(k-m h)^2$              .....(i)

$\Rightarrow \frac{d}{d m}\left(A B^2\right)=2 h-\frac{k}{m} \frac{k}{m^2}-2 h(k-m h)$

$\Rightarrow \frac{d}{d x}(A B)^2=2\left\{\frac{h k}{m^2}-\frac{k^2}{m^3}-h k+m h^2\right\}$

$\Rightarrow \frac{d}{d x}(A B)^2=2\left\{\frac{h k}{m^2}-\frac{k^2}{m^3}-h k+m h^2\right\}$

$\Rightarrow \frac{d}{d x}(A B)^2=\frac{2}{m^3}\left\{m h k-k^2-m^3 h k+m^4 h^2\right\}$

$\Rightarrow \frac{d}{d x}(A B)^2=\frac{2}{m^3}\left\{(m h-k)\left(m^3 h+k\right)\right\}$

∴  $\frac{d}{d m}\left(A B^2\right)=0 \Rightarrow m=\frac{k}{h}$ or, $m=-(k / h)^{1 / 3}$

Since m < 0, therefore $m=-\left(\frac{k}{h}\right)^{1 / 3}$

It can be easily checked that $\frac{d^2}{d m^2}\left(A B^2\right)$ is positive for this value of m. Thus, $A B^2$ is minimum for $m=-(k / h)^{1 / 3}$

Putting $m=-(k / h)^{1 / 3}$ in (i) we get

$A B^2=\left(h^{2 / 3}+k^{2 / 3}\right)^3 A B=\left(h^{2 / 3}+k^{2 / 3}\right)^{3 / 2}$