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I'm a little stuck, will be grateful if someone can help. For the following inequality:

$$\frac{2}{\lambda+X^2}\sum_{t=1}^T(y_t-\hat{y}_t)^2\left(1+\frac{2(y_t-(w,x_t))^2}{\lambda+X^2}\right)\le\|w\|_2^2$$

where $\lambda>0$

I want to simplify such that $\frac{2}{\lambda+X^2}\sum_{t=1}^T(y_t-\hat{y}_t)^2$ is on the left hand side and all the other terms are on the right hand side of the inequality.

The problem I'm having is that I can't move $\left(1+\frac{2(y_t-(w,x_t))^2}{\lambda+X^2}\right)$ on the other side of the inequality due to the summation on the entire expression. Any suggestions?

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Don't know if this is what you want, but we could have $$\text{LHS}=\frac{2}{\lambda+X^2}\sum_{t=1}^T(y_t-\hat{y}_t)^2+\frac{2}{\lambda+X^2}\sum_{t=1}^T\frac{2((y_t-\hat{y}_t)(y_t-(w,x_t)))^2}{\lambda+X^2}$$ so $$\frac{2}{\lambda+X^2}\sum_{t=1}^T(y_t-\hat{y}_t)^2\le\|w\|_2^2-\left(\frac{2}{\lambda+X^2}\right)^2\sum_{t=1}^T((y_t-\hat{y}_t)(y_t-(w,x_t)))^2$$

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  • $\begingroup$ Is it possible to get rid of the term $(y_t-\hat{y}_t)^2$ on the RHS of the inequality? $\endgroup$ – Waqas Mar 1 '18 at 20:38
  • $\begingroup$ Probably not... $\endgroup$ – TheSimpliFire Mar 1 '18 at 20:42

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