If $P(x) = x + \sqrt 2 x^2 + \sqrt{3}x^3+\ldots + \sqrt n x^n$ how can I prove that if $x > 1$ then $$ P(x) \leq \sqrt {\frac{n(n+1)}{2} \frac{x^{2n+1}-1}{x^2-1}}$$
given that $ 1 + 2 + \ldots + n = n(n+1)/2$ and some geometric sum around with $x^2$ terms
As didgogns pointed out, this is an application of the Cauchy-Schwarz inequality. The idea is as follows. $$(\sum_{k=1}^n u_k v_k)^2\leq(\sum_{k=1}^n u_k^2)(\sum_{k=1}^n v_k^2)$$ Substituting $u_k=\sqrt{k}$ and $v_k = x^k$ gives you $$P(x) \leq \sqrt {(\sum_{k=0}^n k)*(\sum_{k=0}^n x^{2n})}$$ And simplifying gives the sums you $$P(x) \leq \sqrt {\frac{n(n+1)}{2} \frac{x^{2n+1}-1}{x^2-1}}$$