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

Question

According to our teacher, we will use the AM-GM Inequality twice

>= is just greater than or equal... There is no greater than or equal sign in the keyboard

Answer 1

Let M be the arithmetic mean of x and y and D be the difference x-M. It can be shown that x=M+D and y=M-D. Thus we can write (M+D)^4 + (M-D)^4 + 8 ≥ 8(M+D)(M-D). Expanding these terms yields (M^4+4M³D+6M²D²+4MD³+D^4) + (M^4-4M³D+6M²D²-4MD³+D^4) + 8 ≥ 8M²-8D². Simplifying somewhat, we get 2M^4+12M²D²+D^4+8 ≥ 8M²-8D². Now assume that our inequality is false - that is, for some x, y, 2M^4+12M²D²+D^4+8 < 8M²-8D². Since M² and D² are always positive or zero, we know that 2M^4+8 ≤ 2M^4+12M²D²+D^4+8 and that 8M²-D² ≤ 8M². Thus we could derive that 2M^4+8 ≤ 2M^4+12M²D²+D^4+8 < 8M²-8D² ≤ 8M² from which it follows that 2M^4+8 < 8M². Since 8M² is positive, we can divide both sides by it without changing the inequality, thus M²/4+1/M² < 1. Let f(M)=M²/4+1/M². The minimum value of this function, if one exists, must occur at a point where its derivative is equal to zero, or where the function is not differentiable. Note that the derivative, f'(M)=M/2-2/M³, is defined everywhere except at M=0, and the function itself isn't defined there either, so f(0) cannot be the minimum. Setting the derivative equal to zero we get: M/2-2/M³=0 ⇒ M^4-4=0 ⇒ M^4=4. The real solutions of this are at ±√2. It is easily shown that both of these points are local minima, as the second derivative, f''(M)=1/2+6/M^4, is always positive. Evaluating f(M) at these points yields f(√2)=1 and f(-√2)=1 Thus, these points are also global minima, and for all real M, f(M)≥1. However earlier we showed that any exception to our inequality would mean f(M) < 1. Therefore, our initial assumption, that there is such an exception, must be false, and it follows that for all real x, y, x^4+y^4+8≥8xy. Q.E.D.

By the way, what's the AM-GM inequality got to do with it?

Answer 2

if x and y are zero, then the left side of the equation is equal to the right. However, if x and y are 1 or any greater number then the left side of the equation will always be larger. For instance, x=2, y=2 then 16+16+8>8*2*2, or 40>32.

Answer 3

divide throughout by 8xy
x^3y/8 + y^3/8 + 1/xy is more than 1 for all values x and y

Answer 4

what does 8> mean? you can't have a > followed by a = unless you did a typo.