Harder 3 unit question! (1 Viewer)

[AbsoluteValue]

The AM-GM inequality states that for any n positive real numbers a_1, a_2, a_3, … a<sub>n:

Prove that:</sub>

Can anyone prove this? It is really tricky but once you get it is easy to solve.

 

[Carrotsticks]

Why is the 'i' there?

 

[Shadowdude]

Where i is the imaginary unit?

 

[AbsoluteValue]

Why is the 'i' there?

Sorry, my bad, it is a factorial I was revising complex numbers today so I'm still into imaginary stuff LOL

 

[Aesytic]

(n+1)/2 >/ root(n*1)
((n-1)+2)/2 >/ root((n-1)*2) ---> (n+1)/2 >/ root(2*(n-1))
((n-2)+3)/2 >/ root((n-2)*3) ---> (n+1)/2 >/ root(3*(n-2))
if we keep doing this, we should eventually get n/2 inequalities similar to those above
multiplying them all together:
[(n+1)^(n/2)]/2^(n/2) >/ root(1*2*3*...*(n-1)*n)
[(n+1)^n/2)]/2^(n/2) >/ root(n!)
(n+1)^(n/2) >/ (2^(n/2)) * root(n!)
squaring both sides,
(n+1)^n >/ 2^n * n!

 

[Trebla]

The result then immediately follows....

 

[AbsoluteValue]

Yeah, that's right, it is really easy but say in the test you don't realise that 1 + 2 + 3 + … + n = n/2 (n+1), this means you can't do the question.

 

[petepopp]

easy