Inspired by Brandon Monsen

$\frac{c}{2}=\sqrt[3]{a^3b+b^3a} \Leftrightarrow \frac{c^3}{8}=(a^2+b^2)ab \Leftrightarrow c=8ab$

Let $\theta$ be the angle between $a$ and $c$ . Then, $\displaystyle c=8c^2\cos(\theta)\sin(\theta) \Leftrightarrow c=\frac{1}{8\cos(\theta)\sin(\theta)} \Leftrightarrow c=\frac{1}{4\sin(2\theta)}$ . It has a maximum $\boxed{\frac{1}{4}}$ when $\theta=\frac{\pi}{4}$ .

Again , bashing!!

Let the vertex containing right angle is $(0,0)$ . Others be $(a,0),(0,b)$ .

Now mid point of hypotenuse is $(a/2,b/2)$ .

Using distance formulae and above given info,

$c=8ab$ .

Or $a^{2} + b^{2}=64(ab)^{2}$ .

Now by $AM-GM$

$a^{2} + b^{2} \geq 2(ab)$ .

Therefore $ab\geq 1/32$ .

Hence $c_{min} = 1/4= H$ .

Therefore answer is $\boxed{4}$ .

Cosine rule for medium m delivers: $m^{2}=\frac{c^{2}}{4}+a^{2}-2a\frac{c}{2}cos\beta=(a^{3}b+b^{3}a)^{\frac{2}{3}}$

With $cos\beta=\frac{a}{c}$ we get $c^{2}=a^{2}+b^{2}=4(a^{3}b+b^{3}a)^{\frac{2}{3}}$

That means we have to minimize $c^{2}=a^{2}+b^{2}$ regarding constraint $a^{2}+b^{2}=4(a^{3}b+b^{3}a)^{\frac{2}{3}}$

This task is solved by the well known Lagrange multiplier method which relults in H=0.25.

Length of median from the vertex $C$ of $\Delta ABC$ with opposite side length $c$ and the other two side lengths $a$ and $b$ is given by

$M = \dfrac 12 \sqrt{2a^2+2b^2-c^2}$

Considering the case of our right triangle, we have $c^2 = a^2 + b^2$ so that

$M = \dfrac 12 \sqrt{a^2+b^2}$

Now,

$\begin{aligned} & \dfrac 12 \sqrt{a^2+b^2} &=& \sqrt[3]{a^3b+b^3a} \\ \implies & \sqrt{a^2+b^2} &=& 2\sqrt[3]{ab} \cdot \sqrt[3]{a^2+b^2} \\ \implies & \sqrt[6]{a^2+b^2} &=& 2\sqrt[3]{ab} \\ \implies & \sqrt[3]{c} &=& 2\sqrt[3]{ab} \\ \implies & c &=& 8ab \end{aligned}$

Using AM-GM inequality, we have

$\begin{aligned} & \dfrac{a^2+b^2}{2} & \ge & \sqrt{a^2b^2} \\ \implies & \dfrac{c^2}{2} & \ge & ab \\ \implies & \dfrac{64a^2b^2}{2} & \ge & ab \\ \implies & 32ab & \ge & 1 \\ \implies & ab & \ge & \dfrac{1}{32} \\ \implies & 8ab & \ge & \dfrac{1}{4} \\ \implies & c & \ge & \dfrac 14 \end{aligned}$

So the minimum value of $c$ is $\mathcal{H} = \dfrac 14$ and thus $16\mathcal{H} = \boxed{4}$ .

$The~medium~=~\frac 1 2 c. ~~~~~~For ~minimum~ c,~legs~ must~ be~ equal.~~So~a=b.~~~~ ~~a^2+b^2=2a^2=c^2.~~~~And~~a=\dfrac c {\sqrt2}.\\ \implies~\frac 1 2 c=\sqrt[3]{a^3b+ab^3}.~~~~~~~\Big(\dfrac c 2 \Big)^3=2a^4=2\Big(\dfrac c {(\sqrt2)^4 }\Big).\\ \therefore~\dfrac{c^3} 8=2*\dfrac{c^4}{4}.~~\implies~c=\dfrac 1 4=H.\\ So~~16*H=16*\frac 1 4=\Large ~~~~\color{#D61F06}{4}.$