# Proving $\prod_{t=1}^{T}\sqrt{1-4x_t^{2}}\leq \exp\left(-2\sum_{t=1}^{T}x_t^{2}\right)$, given $1-x\leq \exp(-x)$

Given that $$1-x\leq \exp{(-x)}$$, prove that

$$\prod_{t=1}^{T}\sqrt{1-4x_t^{2}}\leq \exp\left(-2\sum_{t=1}^{T}x_t^{2}\right)$$

How would one approach this problem. Without the square root I know that

$$1-2\sum_{t=1}^{T}x_t^{2}\leq \exp(-2\sum_{t=1}^{T}x_t^{2})$$

But I'm not sure what to do after that.

For all $$t$$, using the hint, we have $$1-(2x_t)^2\leq \exp\bigg(-(2x_t)^2\bigg)$$ Thus $$1-4x_t^2\leq \exp(-4x^2_t)$$ Take the square root on each side $$\sqrt{1-4x_t^2}\leq \exp(-2x^2_t)$$ Multiplying all these inequalities for $$1\leq t\leq T$$ yields
$$\prod_{t=1}^T\sqrt{1-4x_t^2}\leq \prod_{t=1}^T\exp(-2x^2_t)$$
In other words $$\prod_{t=1}^T\sqrt{1-4x_t^2}\leq \exp(-2 \sum_{t=1}^T x^2_t)$$
• can you explain this $\sqrt{1-4x_t^2}\leq \exp(-2x^2_t)$ How did you get $-2x^2_t$ Shouldnt x^2 become x – user1436508 Nov 8 '18 at 5:01