Question

show that x^3 + y^3 >= (x^2)y + (y^2)x

Answer 1

got any ideas?

Answer 2

x^2+y^2>=0? then amplify by (x-y)? like a thousand years that i don't do this xd

Answer 3

x^2y+xy^2-x^3=<y^3

Answer 4

agreed

Answer 5

what?

Answer 6

that above isn't always true
are there any restrictions on x and y?

Answer 7

well they are intergers

Answer 8

that isn't true for all integers
you can find a counterexample

Answer 9

a counterexample is an example showing the statement is not true for all values within the restriction on the variables

Answer 10

i think it should be solved using arithmetic or geometric median

Answer 11

geometric is best I think

Answer 12

@zupari It is very easy to think of a counterexample.

Answer 13

i don't understand

Answer 14

try x=-1 and y=0
you will see you expression is not true
either the restriction on variables you gave me is wrong
or you are expected to find a counterexample

Answer 15

they are positive intergers

Answer 16

that is a different case
see if you can use (x+y)^3 in helping you with this problem

Answer 17

oh wait they are positive real numbers

Answer 18

tried but no use

Answer 19

hurry. In 6 hours exam starts

Answer 20

x>0 and y>0?
Then (x-y)^2>=0 its true and x+y>=0 its true
So
play with 'em

Answer 21

AM-GM inequality :
\[\large x^2 + y^2 \ge 2xy\]
\[\large x^2 + y^2-xy \ge xy\]

\[\large (x+y)(x^2 + y^2-xy) \ge xy(x+y)\]

\[\large x^3 + y^3 \ge x^2y + y^2 x\]

Answer 22

wow thanks ganeshie8

Answer 23

notice that the above proof was just messing around the identity : \(\large a^3 + b^3 = (a+b)(a^2-ab+b^2)\)

Answer 24

also `x+y>=0 ` was used in line 3
the inequality holds only when you multiply both sides by a nonnegative number

Answer 25

got it