A puzzle
Let u = x - y and v = y - z then the problems is to prove that for u,v>0:
f(u,v) := uv(u + v - 4) + u + v >= 0
To find the minimum of the function lets first calculate the gradient
Df(u,v) = (2uv + v^2 - 3, 2uv + u^2 - 3)
It's only zero when u = v = 1 so the minimum can only be there.
min f(u,v) = f(1,1) = 0 >= 0
q.e.d.
>>7950508
came here to say this
>>7950524
show meeeee
>>7950535
I'm looking for it
>>7950536
wolframalpha.com that shite...lol
>>7950546
nothing good
>>7950536
it's not some bullshit where you formally multiply that shit out and solve a quadratic equation is it
>>7950601
if only i didnt skip pre-calc so fucking much :(
>>7950624
We gonna label (x-y) = a, (y-z) =b
We have (ab)(a+b-4) >= -a -b
Distributing we recieve:
ab^2 +ab^2 -4ab +a +b >= 0
We put them a's and b's together to get:
(b^2 + 1)a + (a^2 +1)b -2ab - 2ab >= 0
We then give each quantity in paratheses there a -2ab and distribute out the respective factor to get:
b*(a^2 -2a +1) + a*(b^2 -2b +1) >=0
We notice that both b and a are positive because of the premise and the quantities in parantheses are squares and thus also nonnegative so the entire left side is non-negative
