2026/085) Prove that sum of cubes of legs of a right angled triangle is less than the cube of its hypotenuse.

Let a b c be sides of right angled triangle where c is the hypotenuse

So $a^2 + b^2 = c^2$

$=>a^2c + b^2c = c^3\cdots(1)$

As $a<c$ we have $a^3<a^2c\cdots(2)$

Also  as $b<c$ we have $b^3<b^2c\cdots(3)$

From (2) and (3) we have

$a^3+b^3 < c(a^2+b^2)$

Using above and (1) we have

$a^3+b^3 < c^3$

Proved