I bet none of you can even do this

i bet none of you can even do this

I bet you can't do your homework either

AM-GM inequality

Well, that's easy.

i bet i can. i will make my trip the answer

wrong

then how

Um, proof by example? Honestly sweetie

there's probably a really neat and obscure way to prove this.

I'd just do proof by exhaustion (or cases).

yeah, if you rewrite it into a better form it seems easy to explain why it's true in each case, for x>y, y>x, x=y, not sure how you could do it all at once though

do a binomial expanion of x and y raised to some even power so you can make the statement that it is greater then or equal to zero?

or maybe you can just make a quadratic in terms of y with it's coefficients being in terms of x and 2, and then show that there can only exist one or zero roots, thus the polynomial is always above the x-axis.

if you have a counter example maybe. i dont think you could show the proof for all the real nums

nope

closest so far

nah its easy. no cases though

will post solution at 1800 est

We have [math]\left(x^2\,-\,y\right)^2 \,=\, x^4 \,-\, 2\,y\,x^2 \,+\, y^2 \,\geqslant\, 0[/math] then [math]2\,y\,x^2 \,\leqslent\, y^2\,+\,x^4[/math], and [math]x^2\,>\,0[/math] hence [math]2\,y \,\leqslent\, \frac{y^2}{x^2}\,+\,x^2[/math].

mfw Veeky Forums legit couldn't do OP's homework

good job senpai

trip was

0

as if there is only one way to prove that..

Easy. I will teach Amerilard brainlets how to solve this middle school exercise :

Turn the expression into 2

I meant y/x^2 + (x^2)/y

>wrong
The geometric mean of [math]2 \frac{y^2}{x^2} [/math] and [math]2 x^2 [/math] is [math]2y [/math] while the arithmetic mean is [math]\frac{y^2}{x^2} + x^2 [/math].

Fuck latex

I'm a brainlet. It should've been (y+x^2)^2 and not 2x