Question

In a right angled triangle `ABC` right angled at `B` prove that `(AB)^3+(BC)^3<(AC)^3`?

Answer 1

In `DeltaABC`, `angle ABC=90^@`.
So `AC` being hypotenuse

`0 < (AB)/(AC)<1and 0< (BC)/(AC)<1`

So

`(AB)^3/(AC)^3<(AB)^2/(AC)^2....[1]`

And

`(BC)^3/(AC)^3<(BC)^2/(AC)^2....[2]`

Adding [1] and [2] we get

`(AB)^3/(AC)^3+(BC)^3/(AC)^3<(AB)^2/(AC)^2+(BC)^2/(AC)^2`

`=>(AB)^3/(AC)^3+(BC)^3/(AC)^3<(AB^2+BC^2)/(AC^2)`

[By Pythagorean theorem

`AB^2+BC^2=AC^2`]

`=>(AB)^3/(AC)^3+(BC)^3/(AC)^3<(AC^2)/(AC^2)=1`

`=>(AB^3+BC^3)/(AC^3)<1`

`=>AB^3+BC^3 < AC^3`

Proved