You can prove a much more general statement by first proving the inequality for weighted means of two elements. I.e. t*a + (1-t)*b >= a^t*b^(1-t), with a,b>0 and t in (0,1). Fix t and b and let f(a) be the difference (which we wanna prove to be nonnegative). We first look at f' and find that f'(a) = t - t*a^(t-1)*b^(1-t). We see that f'(a)=0 at a=b. Furthermore, f' increases so f'(a)
cube ID x^3+ψ^3+z^3-3xψz=1/2(x+ψ+z){(x-ψ)^2+(ψ-z)^2+(z-x)^2}>=0 x^3+ψ^3+z^3>=3xψz if a=qubx b=qubψ z=qubz we proved identity cauchy for three qub means three root
Your proof is easy to understand ❤
What a cool proof ! Bravo !
Thanks!
Ah...finally a great understandable proof for the Am Gm Inequality
😊
Nice proof
Thank you!
You can prove a much more general statement by first proving the inequality for weighted means of two elements. I.e.
t*a + (1-t)*b >= a^t*b^(1-t), with a,b>0 and t in (0,1).
Fix t and b and let f(a) be the difference (which we wanna prove to be nonnegative). We first look at f' and find that f'(a) = t - t*a^(t-1)*b^(1-t). We see that f'(a)=0 at a=b. Furthermore, f' increases so f'(a)
cube ID x^3+ψ^3+z^3-3xψz=1/2(x+ψ+z){(x-ψ)^2+(ψ-z)^2+(z-x)^2}>=0 x^3+ψ^3+z^3>=3xψz if a=qubx b=qubψ z=qubz we proved identity cauchy for three qub means three root
X=Y=Z=3
Bro u could do it easier
(x+y+z)³>=27xyz and then by am gm
Why? 😁