A point on the hypotenuse of a triangle is at distance a and b from the sides of the triangle. Show that the maximum 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<θ<`frac{pi}{2}`
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