$A$ point on the hypotenuse of a triangle is at distances $a$ and $b$ from the sides of the triangle. Show that the minimum length of the hypotenuse is $(a^{\frac{2}{3}}+b^{\frac{2}{3}})^{\frac{3}{2}}$.

(N/A) Let $\triangle ABC$ be a right-angled triangle at $B$. Let $P$ be a point on the hypotenuse $AC$ such that the distance of $P$ from $AB$ is $a$ and from $BC$ is $b$.
Let $\angle C = \theta$. Then $\angle A = 90^{\circ} - \theta$.
In the right-angled triangle formed by the perpendicular from $P$ to $BC$,we have $PC = \frac{b}{\sin \theta} = b \csc \theta$.
In the right-angled triangle formed by the perpendicular from $P$ to $AB$,we have $AP = \frac{a}{\cos \theta} = a \sec \theta$.
The length of the hypotenuse $L = AC = AP + PC = a \sec \theta + b \csc \theta$.
To find the minimum length,we differentiate $L$ with respect to $\theta$:
$\frac{dL}{d\theta} = a \sec \theta \tan \theta - b \csc \theta \cot \theta$.
Setting $\frac{dL}{d\theta} = 0$,we get $a \sec \theta \tan \theta = b \csc \theta \cot \theta$.
$\frac{a \sin \theta}{\cos^2 \theta} = \frac{b \cos \theta}{\sin^2 \theta} \Rightarrow \tan^3 \theta = \frac{b}{a} \Rightarrow \tan \theta = (\frac{b}{a})^{\frac{1}{3}}$.
At this value,$\sin \theta = \frac{b^{\frac{1}{3}}}{(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{1}{2}}}$ and $\cos \theta = \frac{a^{\frac{1}{3}}}{(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{1}{2}}}$.
Substituting these into $L = a \sec \theta + b \csc \theta$:
$L = a \frac{(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{1}{2}}}{a^{\frac{1}{3}}} + b \frac{(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{1}{2}}}{b^{\frac{1}{3}}} = a^{\frac{2}{3}}(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{1}{2}} + b^{\frac{2}{3}}(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{1}{2}}$.
$L = (a^{\frac{2}{3}} + b^{\frac{2}{3}})(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{1}{2}} = (a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{3}{2}}$.
Thus,the minimum length of the hypotenuse is $(a^{\frac{2}{3}} + b^{\frac{2}{3}})^{\frac{3}{2}}$.