Assume that for any real sequence $\{x_{n}\}$, define the sequence $\{y_{n}\}$, such $$y_{1}=x_{1},y_{n+1}=x_{n+1}-\left(\sum_{i=1}^{n}x^2_{i}\right)^{\frac{1}{2}}(n\ge 1)$$ Find the smallest positive number $\lambda$, such for any real sequence $\{x_{n}\}$ and for any positive integer number $m$,have $$\dfrac{1}{m}\sum_{i=1}^{m}x^2_{i}\le\sum_{i=1}^{m}\lambda^{m-i}y^2_{i}$$

and it is said the $$\lambda_{\min}=2$$ and use Mathematical induction can prove it,But I can't.Thank you

This problem is from today china TST,

maybe this inequality is very hard,

Now I can't solve this inequality,and I can't see any this form inequality.

I looking forward to someone can solve this hard problem.

and before china TST inequality awalys from AMM,CRUX Problem,or some paper reslut, I don't know this problem is from AMM,Crux?


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




By definition of $\lambda$, we require that for the above defined $x_i, y_i$,

$$\frac{1}{m}\sum_{i=1}^{m}{x_i^2} \leq \sum_{i=1}^{m}{\lambda^{m-i}y_i^2}$$


$$\frac{1}{m}2^{m-1} \leq \lambda^{m-1}(1)+\sum_{i=2}^{m}{\lambda^{m-i}0^2}=\lambda^{m-1}$$

$$\left(\frac{2}{\lambda}\right)^{m-1} \leq m$$

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, $y_{k+1}=x_{k+1}-\left(\sum_{i=1}^{k}{x_i^2}\right)^{\frac{1}{2}}=b-a$.


\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}


\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 $2$.

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