Indeed, the smallest positive value of $\lambda$ is $2$.
Step 1: $\lambda \geq 2$.
Proof: Let $x_1=1$ and $x_{i+1}=\sqrt{2}^{i-1}$ for $i \geq 1$.
Then
$$\sum_{i=1}^{n}{x_i^2}=1+\sum_{i=2}^{n}{2^{i-2}}=2^{n-1}$$
$$y_1=x_1=1$$
$$y_{n+1}=x_{n+1}-\left(\sum_{i=1}^{n}{x_i^2}\right)^{\frac{1}{2}}=\sqrt{2}^{n-1}-\sqrt{2}^{n-1}=0$$
By definition of $\lambda$, we require that for the above defined $x_i, y_i$,
$$\frac{1}{m}\sum_{i=1}^{m}{x_i^2} \leq \sum_{i=1}^{m}{\lambda^{m-i}y_i^2}$$
Thus
$$\frac{1}{m}2^{m-1} \leq \lambda^{m-1}(1)+\sum_{i=2}^{m}{\lambda^{m-i}0^2}=\lambda^{m-1}$$
$$\left(\frac{2}{\lambda}\right)^{m-1} \leq m$$
For $\lambda<2$, $\frac{2}{\lambda}>1$, so $\left(\frac{2}{\lambda}\right)^{m-1}>m$ for sufficiently large values of $m$, a contradiction.
Thus $\lambda \geq 2$, as desired.
Step 2: $\lambda=2$ works.
Proof: This amounts to proving that
$$\frac{1}{m}\sum_{i=1}^{m}{x_i^2} \leq \sum_{i=1}^{m}{2^{m-i}y_i^2}$$
for any real sequence $\{x_n \}$ and positive integer $m$.
We proceed by induction on $m$.
When $m=1$, the inequality reduces to $x_1^2 \leq y_1^2$, which is trivially true since $y_1=x_1$.
Suppose that the statement holds for $m=k$, i.e. that
$$\frac{1}{k}\sum_{i=1}^{k}{x_i^2} \leq \sum_{i=1}^{k}{2^{k-i}y_i^2}$$
Consider $m=k+1$. Define $a=\left(\sum_{i=1}^{k}{x_i^2}\right)^{\frac{1}{2}}, b=x_{k+1}$. The induction hypothesis then gives
$$\frac{a^2}{k} \leq \sum_{i=1}^{k}{2^{k-i}y_i^2}$$
Now by definition, $y_{k+1}=x_{k+1}-\left(\sum_{i=1}^{k}{x_i^2}\right)^{\frac{1}{2}}=b-a$.
Thus
\begin{align}
\sum_{i=1}^{k+1}{2^{k+1-i}y_i^2}& =y_{k+1}^2+2\sum_{i=1}^{k}{2^{k-i}y_i^2} \\
& \geq y_{k+1}^2+\frac{2a^2}{k} \; \text{by the induction hypothesis} \\
& =(b-a)^2+\frac{2a^2}{k} \\
& =b^2+(1+\frac{2}{k})a^2-2ab \\
\end{align}
Thus
\begin{align}
\sum_{i=1}^{k+1}{2^{k+1-i}y_i^2}-\frac{1}{k+1}\sum_{i=1}^{k+1}{x_i^2} & \geq (b^2+(1+\frac{2}{k})a^2-2ab)-\frac{a^2+b^2}{k+1} \\
& =\frac{k}{k+1}b^2+\left(\frac{k}{k+1}+\frac{2}{k}\right)a^2-2ab \\
& \geq \frac{k}{k+1}b^2+\left(\frac{k-1}{k}+\frac{2}{k}\right)a^2-2ab \; \text{since} \, \frac{k}{k+1} \geq \frac{k-1}{k}\\
& =\frac{k}{k+1}b^2+\frac{k+1}{k}a^2-2ab \\
& = \left( \sqrt{\frac{k}{k+1}}b-\sqrt{\frac{k+1}{k}}a\right)^2\\
& \geq 0
\end{align}
Thus the statement holds for $m=k+1$, so we are done by induction.
In conclusion, the smallest positive value of $\lambda$ is $2$.
