Let $x'$ be the larger root of $x^2 - 2x\sqrt{M} - 1 = 0$ Take $c = x' + 1$ Clearly $c^2 - 2c\sqrt{M} - 1 > 0 $ Now let $M' = c\sqrt{M}$ We will show by induction that $M'$ satisfies the conditions Clearly its true for $n = 1$ Now assume its true for $n = 1,2,...,k$ Then it suffices to show that $\frac{M'}{\sqrt{M}}\sqrt{a_1^2 + a_2^2 ... + a_k^2} > a_1+a_2...+a_k$ $\implies c^2(a_1^2 + a_2^2...+a_k^2) > (a_1 + a_2...+a_k)^2$ Now we will show by induction that $ c^2(a_1^2 + a _2^2 ...+a_m^2) > (a_1 + a_2...+a_m)^2$ is true for $m = 1,2,...k$ Clearly it is true for $ m = 1$ Then $(a_1+a_2...+a_m)^2 = (a_1+a_2...+a_{m-1})^2 + 2a_m(a_1 + a_2...+a_{m-1}) + a_m^2 < c^2(a_1^2 + a _2^2...+a_{m-1}^2) + 2a_m(M'a_m) + a_m^2 = c^2(a_1^2 + a_2^2...+a_{m-1}^2) + (2c\sqrt{M}+1)a_m^2 < c^2(a_1^2 + a_2^2...+a_m^2)$ And thus we are done

Lemma. If $n > m$, then $\frac{Ma_n^2}{a_m^2} > \left(1+\frac{1}{M}\right)^{n-m-1}$. Proof. We will proceed by induction on $n-m$. If $n-m = 1$, then $$ \frac{Ma_{m+1}^2}{a_m^2} > \frac{a_1^2 + \cdots + a_m^2}{a_m^2}\geq\frac{a_m^2}{a_m^2} = 1. $$If $n-m > 1$, then $$ \frac{Ma_n^2}{a_m^2} > \frac{a_1^2 + \cdots + a_{n-1}^2}{a_m^2}\geq\frac{a_m^2 + \cdots + a_{n-1}^2}{a_m^2} = 1 + \frac{1}{M}\left(\frac{Ma_{m+1}^2}{a_m^2} + \cdots + \frac{Ma_{n-1}^2}{a_m^2}\right) $$Applying induction hypotheses to each term in the parenthesis, $$ > 1 + \frac{1}{M}\left(\left(1+\frac{1}{M}\right)^0 + \cdots + \left(1+\frac{1}{M}\right)^{n-m-2}\right) = \left(1+\frac{1}{M}\right)^{n-m-1}.\square $$ It follows from the lemma that, for $k=1,2,\ldots,n$, $a_k < \sqrt{M}\left(1+\frac{1}{M}\right)^{-\frac{n-k}{2}}a_{n+1}$, so $$ a_1 + a_2 + \cdots + a_n < \sqrt{M}a_{n+1}\left(\left(1+\frac{1}{M}\right)^{-\frac{0}{2}} + \left(1+\frac{1}{M}\right)^{-\frac{1}{2}} + \cdots + \left(1+\frac{1}{M}\right)^{-\frac{n}{2}}\right) < M'a_{n+1} $$where $$ M' = \sqrt{M}\left(\left(1+\frac{1}{M}\right)^{-\frac{0}{2}} + \left(1+\frac{1}{M}\right)^{-\frac{1}{2}} + \cdots\right) = (M+1)\sqrt{M} + M\sqrt{M+1}. $$