\\ S_n = \sqrt{\frac{\sum_{k=1}^{n} k^3 }{9}} = \frac{\sqrt {\sum_{k=1}^{n} k^3} }{3} $ but one cool property of the Triangular Numbers is that if we define the nth Triangular Number to be $ T_n $, then $ \sum_{k=1}^{n} k^3 = T_n ^2 $ so what we actually have there is $ S_n = \frac{T_n}{3}...
But it is probably one of the first inequalities you are taught to prove in Harder Extension 1.
Either way, I like asianese's method more because he proved both directions of the statement, having the 'iff', whereas yours only proves one.
I think the confusion came from the fact that suppose we have ma=kv^2, then k can be considered to be mpv^2, where p is some constant, in which case we have a=pv^2. It's just notation, which should be specified in the question.
I'm seeing two contradictory statements. If you have to prove it, then no memorising is required. Also, you used the Cosine Rule. If anything, that is more 'memorising' than asianese's inequality.
3 solutions.
Solution #1: 0 to pi/6
Solution #2: 2pi/3 to 5pi/6
Solution #3: -pi/2 to -2pi/3
All of which are acquired similarly to general solutions.
Interesting note: These loci are used to generate diagrams such as the Radiation Symbol, due to their cyclic nature.
\\ $Find all positive integers $ n $ for which the quadratic equation$ \\\\ a_{n+1}x^2-2x\sqrt{\sum_{i=1}^{n+1}a_i ^2} + \sum_{i=1}^{n} a_i=0 \\\\ $has real roots $ \forall a_i \in \mathbb{R} $ where $ i \in [1,n+1]
I've been told that the notation might be hard to understand for those who haven't seen it before.
(x_1)(x_1+x_2)(x_1+x_2+x_3)(x_1+x_2+x_3+x_4)...(x_1+x_2+x_3+x_4+x_5+...+x_n)