easy by using induction

we prove this ineq when λ=2. m=1 is trival. m -> m+1 : sum(i,1,m+1,y^2*2^(m+1-i))=2*sum(i,1,m,y^2*2^(m-i))+y[m+1]^2≥ (2/m)(sum(i,1,m,x^2))+y[m+1]^2 let k=sqrt(sum(i,1,m,x^2)),x=x[m+1],then the rest is to prove 2k^2/m+(x-k)^2≥(x^2+k^2)/(m+1) by CH-ineq ,(k^2/m+(x-k)^2)(m+1)≥x^2 so 2k^2/m+(x-k)^2≥x^2/(m+1)+k^2/m≥(x^2+k^2)/(m+1) choose x[1]=1,x[i+1]=2^(i-1) for all positive integer i,then y[1]=1,y[i+1]=0, if λ<2,2^m/m>λ^(m-1) for m to be sufficiently large hence λ[min]=2

A small clarification on kite_in_rainbow clever solution: We substitute $x_1^2 = 1$ and $x_k^2= 2^{k-2}$, for $2 \leq k \leq n$, i.e. $x_1^2 = 1$, $x_2^2=1$, $x_3^2=2$, $x_4^2=4, \ldots$ . Then $x_{k+1}^2 = \sum_{i=1}^{k} x_i^2$ and $y_{k+1}= 0$ for all $k \in [1,n]$. Hence, in this case: $\lambda^{n-1}\geq \frac{1}{n} 2^{n-1}$, yielding in the limit $\lambda \geq 2$. Then we need induction on $n$ to finish the proof.

Let $ x_1 = 1 $ and $ x_n = \sqrt{2^{n - 2}} $ for all integers $ n \ge 2. $ Then $ y_n = 0 $ for all integers $ n \ge 2. $ The inequality implies that $ \frac{2^{m - 1}}{m} \le \lambda^{m - 1} $ for all $ m \in \mathbb{N} $ which by letting $ m $ grow arbitrarily large forces $ \lambda \ge 2. $ I claim that the inequality holds when $ \lambda = 2. $ We proceed by induction on $ m, $ where the base case of $ m = 1 $ trivially holds. Assume for some $ k \in \mathbb{N} $ that $ \frac{1}{k}\sum_{i = 1}^{k}x_i^2 \le \sum_{i = 1}^{k}2^{k - i}y_i^2 $ for any suitable sequence $ x_1, x_2, \dots x_k. $ Let $ z = \sqrt{x_1^2 + x_2^2 + \dots + x_k^2}. $ Then $ \sum_{i = 1}^{k + 1}2^{k + 1 - i}y_i^2 = 2\sum_{i = 1}^{k}2^{k - i}y_i^2 + y_{k + 1}^2 \le \frac{2z^2}{k} + (x_{k + 1} - z)^2 $ so it suffices to show that $ \frac{2z^2}{k} + (x_{k + 1} - z)^2 \ge \frac{z^2 + x_{k + 1}^2}{k + 1}. $ But since $ \frac{z^2}{k} > \frac{z^2}{k + 1} $ it suffices to show that $ \left(\frac{z^2}{k} + (x_{k + 1} - z)^2\right)(k + 1) \ge x_{k + 1}^2 $ which is trivial by Cauchy-Schwarz (with casework on the sign of $ x_{k + 1} - z $). Therefore the answer is $ \boxed{\lambda = 2} $