# Given that x > 0, y > 0 and x^2 + y^3 > x^3 + y^4, prove that x^3+y^3<x^2+y^2<x+y<2

### 2 Answers | Add Yours

It is given that x > 0, y > 0 and x^2+y^3>x^3+y^4.

x^2+y^3>x^3+y^4

=> y^3 - y^4 > x^3 - x^2

=> y^3(1 - y) > x^2(x - 1)

As y > 0 and x > 0, y^3 > 0 and x^2 > 0

=> 1 - y > x - 1 ...(1)

If x^3 + y^3 < x^2 + y^2

=> x^3 - x^2 < y^2 - y^3

=> x^2(x - 1) < y^2(1 - y) ...(2)

As x^2 and y^2 are positive and from (1), 1 - y > x - 1, (2) is true.

Again from (1), 1 - y > x - 1

=> x + y < 2 ...(3)

From (2) and (3) it is proved that x^3 + y^3 < x^2 + y^2 < x + y < 2.

**The relation x^3 + y^3 < x^2 + y^2 < x + y < 2. is proved given that x > 0, y > 0 and x^2 + y^3 > x^3 + y^4**

We have given that

`x>0,y>0` and `x^2+y^3>x^3+y^4`

`So`

`x^2-x^3>y^4-y^3`

`x^2(1-x)>y^3(y-1)`

`y-1>0 ,`

`Then`

`(x^2(1-x))/(y^3(y-1))>1`

`x^2,y^3,(y-1)>0 =>(1-x)>0`

`Thus`

`y-1>0 `

`y>1`

`1-x>0`

`1>x`

`=> y>x`

`1-x>y-1`

`2>x+y ---------------(i)`

`x-1<1-y and x<y`

`thereforex(x-1)<y(1-y)`

`x^2-x<y-y^2`

`x^2+y^2<x+y------------------(ii)`

`x<y therefore x^2<y^2`

`so`

`x^2(x-1)<y^2(1-y)`

`x^3-x^2<y^2-y^3`

`x^3+y^3<x^2+y^2 -----------------(iii)`

`Thus,(i) ,(ii), (iii)`

`x^3+y^3<x^2+y^2<x+y<2`