A point on the hypotenuse of a right triangle is at distances a and b from the sides of the triangle. Show that the minimum length of the hypotenuse is `(a^(2/3)+b^(2/3))^(3/2)`

Explanation

Let P be a point on the hypotenuse AC of right `△ABC`
such that `PL ⊥ AB=a `
`PM⊥BC=b`
Let `∠APL=∠ACB=θ `
`AP=asecθ` and `PC=bcosecθ`
Let `l` be the length of the hypotenuse, then `l=AP+PC`
` ⇒a secθ+b cosecθ`, `0<θltfrac{pi}{2}`
Now differentiate `l `with respect to `θ`, we get `frac{dl}{dθ}=asecθ.tanθ−bcosecθ.cotθ`
For maxima and minima `frac{dl}{dθ}=0`
`implies asecθ.tanθ=bcosecθ.cotθ` `implies frac{b}{a}=tan^3 θ`
`implies tanθ=frac{b}{a}^(frac{1}{3})`
`frac{d^2l}{dθ^2}=asecθ(sec^2θ+tan^2θ)+bcosecθ×bcosecθ+cot^2θ`
Since 0<θ<`frac{pi}{2}` so all t ratios of θ are positive. Also a>0 and b<0. `therefore frac{d^2l}{dθ^2}` is positive
⇒ l is least when tanθ=`frac{b}{a}^(frac{1}{3})`
Least value of l =asecθ+bcosecθ `=(a^frac{2}{3}+b^frac{2}{3})^frac{3}{2}`
Hence Proved