Given x and y as member of real number. Could you please show me or prove that x^4+y^4+8>= 8xy?

Question

Please use the Am-GM inequality

Answer 1

x^4+y^4+8>= 8xy?

I redid the proof using the Am-Gm inequality.
Using Am-Gm
let x^4=(x²)² & y^4=(y²)²
((x²)²+ (y²)²)/2 >= sqrt{(x²*y²)²} =>

((x²)²+ (y²)²)>= 2*x²*y²

((x²)²+ (y²)²)+8>= 2*x²*y²+8
Using Am-Gm again on 2*x²*y²+8 =>
(2*x²*y²+8)/2>=sqrt(2*x²*y²*8) =>
(2*x²*y²+8)>= 2*sqrt(16*x²*y²) =>
(2*x²*y²+8)>= 2*4*x*y=8xy
Since x^4+y^4+8>= (2*x²*y²+8) &
(2*x²*y²+8)>=8*x*y
then x^4+y^4+8>= 8xy.
end of proof using the Am-Gm twice

Answer 2

Change variables:
x = u+v
y = u-v
We must show that
(u+v)^4 + (u-v)^4 + 8 >= 8(u+v)(u-v)
After expanding this out and cancelling etc
This is the same as
u^4 - 4u^2 + 4 + v^4 + v^2 + 6u^2v^2 >= 0
or
(u^2-2)^2 + v^4 + v^2 +6(u^2)(v^2) >= 0

The last inequality is true because all
terms are squares. I think this is right.
I may have made an algebra error though.

I didn't use whatever the AmGM inequality is, sorry.
Do I get credit for this or do you :)?

Answer 3

From the AM-GM inequality,
(x^4+y^4)/2 >=sqrt(x^4y^4)=x^2y^2
x^4+y^4>=2x^2y^2

If the quantity 2x^2y^2 is greater than or equal to 8xy-8, then the statement is true since:
x^4+y^4>=8xy-8 would have to be true.

So all we need to prove is that 2x^2y^2>=8xy-8. Subtract 8xy-8 from both sides and divide by 2 results in:
x^2y^2 - 4xy+4>=0
(xy-2)^2>=0, which is true since the square of any real number is greater than or equal to 0.

Answer 4

Sorry, I think Am-GM is some abbreviation your teacher or book uses that the rest of us might not recognize. (The A-minor-G-Major inequality, maybe? ;-) )

Answer 5

for x=1 and y=1 the sentance is true

Answer 6

I am very bad with Maths.